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RIEMANN-HILBERT APPROACH AND DOUBLE-POLE SOLUTIONS FOR THE

THIRD-ORDER FLOW EQUATION OF THE KAUP-NEWELL SYSTEM WITH

ZERO/NONZERO BOUNDARY CONDITIONS

JUNCAI PU AND YONG CHEN∗

Abstract. The Kaup-Newell (KN) system is a significant integrable system, which reveals many physical phenomenaand natural laws. Especially, since the solutions of higher-order KN systems are more complex and various than thatof second-order flow KN systems which contain the derivative nonlinear Schrodinger equation (DNLS), Chen-Lee-Liuequation, Gerdjikov-Ivanov equation and so on, the research on it will become more and more important. In thispaper, we investigate the third-order flow equation of Kaup-Newell system (TOFKN) with zero boundary conditions(ZBCs) and non-zero boundary conditions (NZBCs) by using the Riemann-Hilbert (RH) approach based on the inversescattering transformation. Starting from Lax pairs of the TOFKN equation, the direct scattering problem constructsthe analyticity, symmetries, asymptotic behaviors and discrete spectral properties for the Jost solutions, modified Jostsolutions and scattering matrix. The inverse scattering problems are solved by establishing the matrix RH problem,and the reconstruction formula of potential, trace formula and theta condition are also derived correspondingly. Thereflectionless potentials with double poles are exhibited explicitly by means of determinants. Specifically, vivid plots anddynamics analysis which contain double-pole soliton solutions for the ZBCs and double-pole dark-bright soliton solution,breather-breather solutions and breather-breather-dark-bright interaction solutions for the NZBCs, are demonstratedin details. Furthermore, compared with the most classical second-order flow KN system which also be called the DNLSequation, the third-order dispersion and quintic nonlinear term of the KN system change the trajectory and velocity ofsolutions.

1. Introduction

For decades, nonlinear integrable systems, which are used to describe complex natural phenomena in the real world,have always been the research hotspots in the field of mathematics and physics [1,2]. Recently, since abundant solutionsfor nonlinear integrable systems are one of the most significant characteristics to reveal complex natural phenomena,the study of the integrability and exact solutions for nonlinear integrable systems have been paid more and moreattention in optical fiber, fluid dynamics, plasma physics, machine learning and others fields [3–7]. General speaking,it is difficult to find the exact solutions of complex nonlinear integrable systems, with the development of soliton theoryin recent decades, some effective methods for solving nonlinear integrable systems have been established, such as inversescattering transformation [8], Hirota bilinear method [9], Darboux transformation [10], Backlund transformation [11],symmetry reduction [12] and deep learning method [7].

Among these methods mentioned above, inverse scattering transformation (IST) is one of the most importantand basic theories for solving nonlinear integrable systems. In 1967, Gardner, Greene, Kruskal and Miura (GGKM)discovered the classical IST to solve the initial value problem for the Korteweg-de Vries (KdV) equation with laxpairs [8]. After that, many researchers try to extend this method to other nonlinear integrable systems which possessso-called lax pairs [1]. Zakharov and Shabat studied the IST of nonlinear Schrodinger (NLS) equation in 1972 [13].Later, Ablowitz, Kaup, Newell and Segur (AKNS) proposed a new class of integrable systems, called AKNS systems,and established a general framework for their ISTs in 1973 [14,15]. Subsequently, many classical nonlinear integrablesystems are proved to be solvable by IST [16–19]. However, it is found that the solving process of the classical IST fornonlinear integrable systems with second order spectral problems which was based on the Gel’fand-Levitan-Marchenkointegral equations is complicated and tedious, and it is difficult to solve the original IST for the higher order spectralproblems of nonlinear integrable systems without the Gel’fand-Levitan-Marchenko theory [20]. Later on, a Riemann-Hilbert (RH) approach was developed which streamlines and simplifies the IST considerably by Zakharov and hiscollaborators [21]. Since 1980s, the RH approach has been applied to nonlinear integrable systems as a more generalmethod than the classical IST. In 1993, Deift and Zhou presented a new and general steepest descent approach toanalyzing the asymptotics of oscillatory RH problems, and such problems will arise in the process of evaluating thelong-time behavior of nonlinear wave equations solvable by the RH method [22]. Recently, the RH approach hasbecame a powerful tool for constructing IST to obtain abundant solutions of nonlinear integrable systems and dealingwith the long time asymptotic behavior of solutions by analysing RH problem. The multisoliton solutions of someimportant nonlinear integrable systems can be derived by solving a particular RH problem under the reflectionlesscases [23–29]. On the other hand, the RH approach has also been utilized to study the initial boundary value problemsand the long-time asymptotic behavior for many nonlinear integrable systems [22, 30–33].

The Kaup-Newell (KN) systems, which are more complex than AKNS system, are of great significance in mathe-matical physics [34]. The derivative nonlinear Schrodinger equation (DNLS) is the most classical KN system, whichwas first derived from Alfven wave propagation in plasma by Mio et al. in 1976, and well described the propagation

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http://arxiv.org/abs/2105.06098v2

2 JUNCAI PU AND YONG CHEN∗

of small amplitude nonlinear Alfven wave in low plasma [36]. In Ref. [34], Kaup and Newell obtained the one-solitonsolution and the infinity of conservation laws for the DNLS via using inverse scattering technique. Furthermore, otherimportant KN systems have been widely studied by means of IST and RH approach, such as the Chen-Lee-Liu equa-tion [37], the Kundu-type equation [38] and the Gerdjikov-Ivanov equation [39]. However, as far as we know, theseIST works on these nonlinear integrable systems with ZBCs/NZBCs mainly focus on the case that all discrete spectraare simple. A new research topic is to obtain explicit double-pole solutions for nonlinear integrable systems withZBCs/NZBCs by utilizing RH method based on IST, and there are few related works on this aspect at present [40–42].It is worth mentioning that Pichler presented IST for the focusing nonlinear Schrodinger equation with non-zeroboundary conditions at infinity and double zeros of the analytic scattering coefficients [43]. In addition, the IST anddouble-pole solutions for some important nonlinear integrable systems have been investigated [40,42]. Recently, Zhangand Yan proposed a rigorous theory of the ISTs for the DNLS equation with both ZBCs and NZBCs at infinity anddouble zeros of analytical scattering coefficients, provided a significant foundation for the IST for KN systems [44].In this paper, we consider the third-order flow equation of the KN system (TOFKN equation) which contains thethird-order dispersion and quintic nonlinear terms. In 1999, Imai firstly proposed the coupled TOFKN equation [45]and pointed out that it is an integrable system, the coupled TOFKN equation as shown below

qt =a64ε2

(qxxx −

3(qrqx)xε

+3(q3r2)x

2ε2

),

rt =a64ε2

(rxxx +

3(qrrx)xε

+3(q2r3)x

2ε2

),

where q = q(x, t), r = r(x, t).By imposing the condition r(x, t) = −q(x, t)∗ (superscript “∗” denotes complex conjugation) and taking appropriate

parameters ε = i, a6 = 4 to the coupled TOFKN equation, the coupled TOFKN equation is reduced and obtained thegeneral form of the TOFKN equation:

qt + qxxx − 3i(|q|2qx)x −3

2(|q|4q)x = 0, (1.1)

where |q|2 = q(x, t)q∗(x, t). To the best of our knowledge, there are few studies on Eq. (1.1), and the inverse scatteringtransformation for equation (1.1) has not been investigated by utilizing RH method. Lin et al. studied this equationand derived different types of solutions which contain solitons, positons, breathers and rogue waves by using Darbouxtransformation and extended Darboux transformation for the KN systems [46]. The TOFKN equation is completelyintegrable and associated with the following modified Zakharov-Shabat eigenvalue problem (Lax pairs) [34]:

Ψx = XΨ, (1.2)

Ψt = TΨ, (1.3)

where

X = X(x, t, λ) = λ(iλσ3 +Q),

T = T (x, t, λ) =

(4λ4 +

3

2|q|4 − 2λ2|q|2

)X +

(−2iλ3 + 3iλ|q|2

)σ3Qx − λ2(−qq∗x + qxq

∗)σ3 − λQxx,

it is easy to check that Eq. (1.1) as the integrability condition (or zero-curvature condition) Ψxt = Ψtx(or Xt − Tx +

[X,T ] = 0)of system Eqs. (1.2)-(1.3), Ψ = Ψ(λ;x, t) is a 2 × 2 matrix-valued eigenfunction, λ ∈ C, the potential

matrix Q = Q(x, t) is written as

Q =

(0 q(x, t)

−q∗(x, t) 0

), (1.4)

the σ3 is one of the Pauli’s spin matrices given by

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

). (1.5)

This paper is organized as follows. In Section 2, the IST for the TOFKN equation (1.1) with ZBC at infinity isintroduced and solved for the double zeros of analytically scattering coefficients by means of the matrix RH problem.The determinant form of explicit N-double-pole solutions have been presented, and the relative plots at N = 1, 2 havebeen given out in detail. Furthermore, compared with the DNLS in the case of N = 1, and the correlation analysesare illustrated in detail. In Section 3, the IST for the TOFKN equation (1.1) with NZBC at infinity is introduced andsolved for the double zeros of analytically scattering coefficients. The IST for NZBC at infinity is more complicatedthan the case of ZBCs since more symmetries and multivalued functions, thus we map the original spectral parameterλ into single-valued parameter z by introducing an appropriate two-sheeted Riemann surface. As a result, we presentan explicit expression for the N-double-pole solutions for the case of NZBCs by using determinants, and exhibit vivid

RH APPROACH AND DOUBLE-POLE SOLUTIONS FOR THE TOFKN WITH ZERO/NONZERO BOUNDARY CONDITIONS 3

plots of three different types of double-pole solutions in detail. Moreover, some comparison made between the double-pole dark-bright soliton solution of the TOFKN equation and the DNLS equation, and the correlation analyses areexplained detailly. Summary and conclusion are given out in last section.

2. The IST with ZBCS and Double Poles Solution

In this section, we will seek the N-double-pole solution q(x, t) for the TOFKN equation (1.1) with ZBCs at infinityas follow

q(x, t) ∼ 0, as x→ ±∞. (2.1)

In the following subsection, we will present the IST which contain the direct scattering and the inverse problem forEq. (1.1) with ZBCs by RH method respectively.

2.1. The Direct Scattering with ZBCs.

2.1.1. Jost Solution, Analyticity and Continuity.

Considering the asymptotic scattering spectrum problem (x → ∞) of the modified Zakharov-Shabat eigenvalueproblem

Ψx = X0Ψ, (2.2)

Ψt = T0Ψ, (2.3)

where X0 = iλ2σ3 and T0 = 4λ4X0 = 4iλ6σ3, one can obtain the fundamental matrix solution Ψbg(λ;x, t) of Eqs.(2.2)-(2.3)

Ψbg(λ;x, t) = eiθ(λ;x,t)σ3 , θ(λ;x, t) = λ2(x+ 4λ4t).

Let Σ := R ∪ iR. Then, the Jost solutions ψ±(λ;x, t) can be derived as below

ψ±(λ;x, t) ∼ eiθ(λ;x,t)σ3 , λ ∈ Σ, as x→ ±∞. (2.4)

In order to obtain the modified Jost solution µ±(λ;x, t), we consider a transformation of the form

µ±(λ;x, t) = ψ±(λ;x, t)e−iθ(λ;x,t)σ3 , (2.5)

it is evident that

µ±(λ;x, t) ∼ I, as x→ ±∞,

where I is 2× 2 identity matrix. According to the modified Zakharov-Shabat eigenvalue problem Eqs. (1.2)-(1.3) andtransformation Eq. (2.5), one can obtain µ±(λ;x, t) satisfy the following equivalent Lax pair:

µ±,x(λ;x, t) + iλ2[µ±(λ;x, t), σ3] = λQ(x, t)µ±(λ;x, t), (2.6)

µ±,t(λ;x, t) + 4iλ6[µ±(λ;x, t), σ3] = [T (λ;x, t)− T0]µ±(λ;x, t), (2.7)

where the Lie bracket [L1, L2] = L1L2 − L2L1. Eqs. (2.6) and (2.7) can be written in full derivative form

d(e−iθ(λ;x,t)σ3µ±(λ;x, t)) = e−iθ(λ;x,t)σ3{[λQ(x, t)dx + [T (λ;x, t)− T0]dt

]µ±(λ;x, t)

}, (2.8)

and µ±(λ;x, t) satisfy the Volterra integral equations

µ±(λ;x, t) = I +

∫ x

±∞

eiλ2(x−y)σ3(λQ(y, t)µ±(λ; y, t))dy, (2.9)

where eχσ3E := eχσ3Ee−χσ3 with E being a 2× 2 matrix. Let D± := {λ ∈ C| ±Re(λ)Im(λ) > 0}, as shown in Fig. 1.Furthermore, one has the following proposition.

Proposition 1. Suppose that q(x, t) ∈ L1(R) and ψ±i(λ;x, t) (µ±i(λ;x, t)) represent the ith column of ψ±(λ;x, t)(µ±(λ;x, t)). Then, the Jost solutions ψ±(λ;x, t)

(modified Jost solutions µ±(λ;x, t)

)possess the properties:

• Eq. (1.2) has the unique Jost solutions ψ±(λ;x, t)(modified Jost solutions µ±(λ;x, t)

)satisfying Eq. (2.4) (combination

of Eqs. (2.4)-(2.5)) on Σ.• The column vectors ψ+1(λ;x, t) (µ+1(λ;x, t)) and ψ−2(λ;x, t) (µ−2(λ;x, t)) can be analytically extended to D+ andcontinuously extended to D+ ∪ Σ.• The column vectors ψ+2(λ;x, t) (µ+2(λ;x, t)) and ψ−1(λ;x, t) (µ−1(λ;x, t)) can be analytically extended to D− andcontinuously extended to D− ∪ Σ.


Reλ

Imλ

0

λn

λ∗n

−λ∗n

−λn

Figure 1. (Color online) Distribution of the discrete spectrum and jumping curves for the Riemann-Hilbert problem on complex λ-plane. Region D+ =

{λ ∈ C

∣∣ReλImλ > 0}

(yellow region), region

D− ={λ ∈ C

∣∣ReλImλ < 0}(white region).

Proof. We define modified Jost solutions µ± = (µ±1, µ±2) =

(µ±11 µ±21

µ±12 µ±22

), in which µ±1 and µ±2 is the first and

second columns of µ±, respectively. Then, taking µ− as an example, Eq. (2.9) can be rewritten as(µ−11 µ−21

µ−12 µ−22

)=

(1 00 1

)+ λ

∫ x

±∞

(qµ−12 qµ−22e

2iλ2(x−y)

−q∗µ−11e−2iλ2(x−y) −q∗µ−21

)dy,

note that µ−1 =

(10

)+ λ

∫ x±∞

(qµ−12

−q∗µ−11e−2iλ2(x−y)

)dy, where e−2iλ2(x−y) = e−2i(x−y)Re(λ2)e2(x−y)Im(λ2) =

e−2i(x−y)Re(λ2)e4(x−y)Re(λ)Im(λ). Since x − y > 0 and Re(λ)Im(λ) < 0, it indicate that the first column of µ− isanalytically extended to D−. In addition, since Im(λ2) = 0 when λ ∈ Σ, it demonstrate that the first column of µ− iscontinuously extended to D−∪Σ. In the same way, we also obtain the analyticity and continuity of µ−2, µ+1 and µ+2.Similarly, the analyticity and continuity for the Jost solutions ψ±(λ;x, t) can be simply shown from ones of µ±(λ;x, t)and the relation (2.5).

�

Proposition 2. The Jost solutions ψ±(λ;x, t) satisfy both parts of the modified Zakharov-Shabat eigenvalue problem(1.2)-(1.3) simultaneously.

Proof. The Liouville’s formula leads to

det(ψ±(λ;x, t)) = limx→±∞

det(ψ±(λ;x, t)) = limx→±∞

det(µ±(λ;x, t)) = 1,

that show ψ±(λ;x, t) are the fundamental matrix solutions on Σ. According to the zero-curvature condition Xt−Tx+[X,T ] = 0, one can obtain that ψ±,t(λ;x, t)− Tψ±(λ;x, t) also solve the x-part (1.2), that is

(ψ±,t(λ;x, t) − Tψ±(λ;x, t))x = X(ψ±,t(λ;x, t) − Tψ±(λ;x, t)),

ψ±,tx(λ;x, t) − Txψ±(λ;x, t)− Tψ±,x(λ;x, t) = Xψ±,t(λ;x, t) −XTψ±(λ;x, t),

(Xψ±(λ;x, t))t − Txψ±(λ;x, t) − TXψ±(λ;x, t) = Xψ±,t(λ;x, t)−XTψ±(λ;x, t),

Xtψ±(λ;x, t) +Xψ±,t(λ;x, t) − Txψ±(λ;x, t) − TXψ±(λ;x, t) = Xψ±,t(λ;x, t) −XTψ±(λ;x, t),

(Xt − Tx +XT − TX)ψ±(λ;x, t) = 0.

Thus, there exist the two matrices U±(λ; t) such that

ψ±,t(λ;x, t) − Tψ±(λ;x, t) = ψ±(λ;x, t)U±(λ; t), as λ ∈ Σ,

multiplying both sides by e−iθ(λ;x,t)σ3 , we have

ψ±,t(λ;x, t)e−iθ(λ;x,t)σ3 − Tψ±(λ;x, t)e

−iθ(λ;x,t)σ3 = ψ±(λ;x, t)U±(λ; t)e−iθ(λ;x,t)σ3, as λ ∈ Σ,

(ψ±(λ;x, t)e−iθ(λ;x,t)σ3)t + T0ψ±(λ;x, t)e−iθ(λ;x,t)σ3 − Tψ±(λ;x, t)e

−iθ(λ;x,t)σ3 = ψ±(λ;x, t)U±(λ; t)e−iθ(λ;x,t)σ3 , as λ ∈ Σ,

(µ±(λ;x, t))t + T0µ±(λ;x, t)− Tµ±(λ;x, t) = ψ±(λ;x, t)U±(λ; t)e−iθ(λ;x,t)σ3 , as λ ∈ Σ,

and letting x → ±∞, one can find µ±(λ;x, t) ∼ I, T ∼ T0, and U±(λ; t) = 0, that is, ψ±(λ;x, t) also solve the t-part(1.3).


�

2.1.2. Scattering Matrix and Reflection Coefficients.

Since the Jost solutions ψ±(λ;x, t) solve the both parts of the modified Zakharov-Shabat eigenvalue problem (1.2)-(1.3). Therefore, there exist a constant scattering matrix S(λ) = (sij(λ))2×2 independent of x and t to satisfy thelinear relation between ψ+(λ;x, t) and ψ−(λ;x, t), where sij(λ) are called the scattering coefficients, we have

ψ+(λ;x, t) = ψ−(λ;x, t)S(λ), λ ∈ Σ, (2.10)

let ψ+(λ;x, t) = (ψ+1, ψ+2) =

(ψ+11 ψ+12

ψ+21 ψ+22

), ψ−(λ;x, t) = (ψ−1, ψ−2) =

(ψ−11 ψ−12

ψ−21 ψ−22

), substituting into Eq.

(2.10) and using Cramer’s rule, one can obtain

s11(λ) =

det

(ψ+11 ψ−12

ψ+12 ψ−22

)

det(ψ−(λ;x, t))= det(ψ+1(λ;x, t), ψ−2(λ;x, t)),

s12(λ) =

det

(ψ+12 ψ−12

ψ+22 ψ−22

)

det(ψ−(λ;x, t))= det(ψ+2(λ;x, t), ψ−2(λ;x, t)),

s21(λ) =

det

(ψ−11 ψ+11

ψ−21 ψ+21

)

det(ψ−(λ;x, t))= det(ψ−1(λ;x, t), ψ+1(λ;x, t)),

s22(λ) =

det

(ψ−11 ψ+12

ψ−21 ψ+22

)

det(ψ−(λ;x, t))= det(ψ−1(λ;x, t), ψ+2(λ;x, t)),

(2.11)

where det(ψ−(λ;x, t)) = 1.

Proposition 3. Suppose that q(x, t) ∈ L1(R). Then, s11(λ) can be analytically extended to D+ and continuouslyextended to D+ ∪Σ, while s22(λ) can be analytically extended to D− and continuously extended to D− ∪Σ. Moreover,both s12(λ) and s21(λ) are continuous in Σ.

Proof. From proposition (1), we know that ψ+1(λ;x, t) and ψ−2(λ;x, t) can be extended analytically to D+ and con-tinuously extended to D+ ∪Σ. Since s11(λ) = det(ψ+1(λ;x, t), ψ−2(λ;x, t)), one can obtain s11(λ) can be analyticallyextended to D+ and continuously extended to D+ ∪Σ. Similarly, we can prove s22(λ) can be analytically extended toD− and continuously extended to D− ∪ Σ. Due to ψ±1(λ;x, t) and ψ±2(λ;x, t) are continuous in Σ, both s12(λ) ands21(λ) are continuous in Σ. �

Note that it cannot be ruled out that s11(λ) and s22(λ) may exist zeros along Σ. In order to study the RH problemin the inverse process, we focus on the potential without spectral singularity. In general, the reflection coefficientsρ(λ) and ρ(λ) are defined by

ρ(λ) =s21(λ)

s11(λ), ρ(λ) =

s12(λ)

s22(λ). (2.12)

2.1.3. Symmetry Properties.

Proposition 4. X(λ;x, t), T (λ;x, t), Jost solutions, modified Jost solutions, scattering matrix and reflection coeffi-cients have two kinds of symmetry reductions as follow:• The first symmetry reduction

X(λ;x, t) = σ2X(λ∗;x, t)∗σ2, T (λ;x, t) = σ2T (λ∗;x, t)∗σ2,

ψ±(λ;x, t) = σ2ψ±(λ∗;x, t)∗σ2, µ±(λ;x, t) = σ2µ±(λ

∗;x, t)∗σ2,

S(λ) = σ2S(λ∗)∗σ2, ρ(λ) = −ρ(λ∗)∗.

(2.13)

• The second symmetry reduction

X(λ;x, t) = σ1X(−λ∗;x, t)∗σ1, T (λ;x, t) = σ1T (−λ∗;x, t)∗σ1,ψ±(λ;x, t) = σ1ψ±(−λ∗;x, t)∗σ1, µ±(λ;x, t) = σ1µ±(−λ∗;x, t)∗σ1,

S(λ) = σ1S(−λ∗)∗σ1, ρ(λ) = ρ(−λ∗)∗.(2.14)


Proof. First, let’s prove the first symmetry reduction. Since we know the specific forms of the matrices X and T , wecan obtain the symmetry reduction of X and T by direct calculation. In order to obtain the symmetry reduction forJost solutions, in fact, we just need to demonstrate that σ2ψ±(λ

∗;x, t)∗σ2 is also a solution of Eq. (1.2) and admitsthe same asymptotic behavior as the Jost solutions ψ±(λ;x, t).

Considering

ψ±,x(λ;x, t) = X(λ;x, t)ψ±(λ;x, t), (2.15)

replacing λ with λ∗, and taking conjugate both sides simultaneously, we have ψ±,x(λ∗;x, t)∗ = X(λ∗;x, t)∗ψ±(λ

∗;x, t)∗,

substituting X(λ∗;x, t)∗ = σ−12 X(λ;x, t)σ−1

2 into previous formula, then both sides of the obtained formula multiplyingmatrix σ2 on left and right at the same time, one can obtain

σ2ψ±,x(λ∗;x, t)∗σ2 = σ2σ

−12 X(λ;x, t)σ−1

2 ψ±(λ∗;x, t)∗σ2,

⇒σ2ψ±,x(λ∗;x, t)∗σ2 = X(λ;x, t)σ−1

2 (σ−12 σ2)ψ±(λ

∗;x, t)∗σ2,

⇒σ2ψ±,x(λ∗;x, t)∗σ2 = X(λ;x, t)σ−1

2 σ−12 (σ2ψ±(λ

∗;x, t)∗σ2),

where σ−12 =

(0 i−i 0

), so σ−1

2 σ−12 = I, we have

σ2ψ±,x(λ∗;x, t)∗σ2 = X(λ;x, t)(σ2ψ±(λ

∗;x, t)∗σ2). (2.16)

By comparing Eqs. (2.15) and (2.16), we know that σ2ψ±,x(λ∗;x, t)∗σ2 is also a solution of Eq. (2.15), and

ψ±(λ;x, t) = σ2ψ±(λ∗;x, t)∗σ2 because they have the same asymptotic behavior

ψ±(λ;x, t), σ2ψ±(λ∗;x, t)∗σ2 ∼ I, as x→ ±∞.

According to Eq. (2.5) and θ(λ∗)∗ = θ(λ), we can obtain ψ±(λ;x, t) = µ±(λ;x, t)eiθ(λ;x,t)σ3 and ψ±(λ

∗;x, t)∗ =µ±(λ

∗;x, t)∗e−iθ(λ;x,t)σ3 . Substituting into ψ±(λ;x, t) = σ2ψ±(λ∗;x, t)∗σ2, one can derive µ±(λ;x, t)e

iθ(λ;x,t)σ3 =σ2µ±(λ

∗;x, t)∗e−iθ(λ;x,t)σ3σ2 ⇒ µ±(λ;x, t) = σ2µ±(λ∗;x, t)∗e−iθ(λ;x,t)σ3σ2e

−iθ(λ;x,t)σ3 = σ2µ±(λ∗;x, t)∗σ2.

Next, we demonstrate the symmetry reduction of the scattering matrix. We substitute λ = λ∗ into Eq. (2.10)and conjugate both sides of the equation. We have ψ+(λ

∗;x, t)∗ = ψ−(λ∗;x, t)∗S(λ∗)∗, From the symmetry re-

duction of Jost solutions, one can obtain σ−12 ψ+(λ;x, t)σ

−12 = σ−1

2 ψ−(λ;x, t)σ−12 (σ−1

2 σ2)S(λ∗)∗ ⇒ ψ+(λ;x, t) =

ψ−(λ;x, t)σ−12 σ−1

2 (σ2S(λ∗)∗σ2), then

ψ+(λ;x, t) = ψ−(λ;x, t)(σ2S(λ∗)∗σ2). (2.17)

By comparing Eqs. (2.10) and (2.17), we can obtain S(λ) = σ2S(λ∗)∗σ2. According to the symmetry reduction of

scattering matrix, we have(s11(λ) s12(λ)s21(λ) s22(λ)

)=

(0 −ii 0

)(s11(λ

∗)∗ s12(λ∗)∗

s21(λ∗)∗ s22(λ

∗)∗

)(0 −ii 0

)=

(s22(λ

∗)∗ −s21(λ∗)∗−s12(λ∗)∗ s11(λ

∗)∗

),

we then obtain these symmetric relations

s11(λ) = s22(λ∗)∗, s12(λ) = −s21(λ∗)∗, s21(λ) = −s12(λ∗)∗, s22(λ) = s11(λ

∗)∗, (2.18)

Since ρ(λ) = s21(λ)s11(λ)

= −s12(λ∗)∗

s22(λ∗)∗ = −ρ(λ∗)∗, the symmetry reduction of reflection coefficient is ρ(λ) = −ρ(λ∗)∗.Similarly, by repeating the above process, we can prove the second symmetry reduction. This completes the proof.

�

2.1.4. Asymptotic Behaviors.

In order to propose and solve the matrix RH problem for the inverse problem in the following part, the asymptoticbehavior of the modified Jost solution and the scattering matrix must be determined as λ → ∞. The asymptoticbehaviors of the modified Jost solution can be derived by employing the usual Wentzel-Kramers-Brillouin (WKB)expansion.

Proposition 5. The asymptotic behaviors of the modified Jost solutions are as follows:

µ±(λ;x, t) = eiν±(x,t)σ3 +O

(1

λ

), as λ→ ∞, (2.19)

where functions ν±(x, t) read as

ν±(x, t) =1

2

∫ x

±∞

|q(y, t)|2dy. (2.20)


Proof. By utilizing the WKB expansion, we substitute µ±(λ;x, t) =n∑i=0

µ[i]±

(x,t)

λi + O(1λ

)( as λ → ∞) into equivalent

Lax pair (2.6), and compare the coefficients of different powers of λ, we have

O(λ2) : iµ[0]± (x, t)σ3 − iσ3µ

[0]± (x, t) = 0,⇒

(µ[0]± (x, t)

)off= 0, (2.21)

where(µ[0]± (x, t)

)offrepresent the off-diagonal parts of µ

[0]± (x, t).

O(λ) : iµ[1]± (x, t)σ3 − iσ3µ

[1]± (x, t) = Q(x, t)µ

[0]± (x, t),⇒

(µ[1]± (x, t)

)off=i

2σ3Q(x, t)µ

[0]± (x, t), (2.22)

O(1) : µ[0]±,x(x, t) + iµ

[2]± (x, t)σ3 − iσ3µ

[2]± (x, t) = Q(x, t)µ

[1]± (x, t),⇒ µ

[0]± (x, t) = Cdiageiν±(x,t)σ3 , (2.23)

where ν±(x, t) = 12

∫ x±∞ |q(y, t)|2dy, µ±(λ;x, t) = Cdiageiν±(x,t)σ3 + O

(1λ

), as λ → ∞. When x → ±∞, we have

ν±(x, t) → 0, µ±(λ;x, t) → I, one can obtain Cdiag = I, and deduce the asymptotic behavior as λ→ ∞.�

Proposition 6. The asymptotic behavior for the scattering matrix is as follows:

S(λ) = e−νσ3 +O

(1

λ

), as λ→ ∞, (2.24)

where the constant ν reads as

ν =1

2

∫ +∞

−∞

|q(y, t)|2dy. (2.25)

Proof. From the relationship among scattering matrix, Jost solution and modified Jost solution, we have

µ+(λ;x, t)eiθ(λ;x,t)σ3 = µ−(λ;x, t)e

iθ(λ;x,t)σ3S(λ).

According to proposition (5), when λ→ ∞, µ±(λ;x, t) = eiν±(x,t)σ3 +O(1λ

), we have

[eiν+(x,t)σ3 +O

(1

λ

)]eiθ(λ;x,t)σ3 =

[eiν−(x,t)σ3 +O

(1

λ

)]eiθ(λ;x,t)σ3S(λ),

then both sides of the equation pre-multiply e−i[ν−(x,t)+θ(λ;x,t)]σ3 at the same time, one can obtain

S(λ) = e−i[ν−(x,t)−ν+(x,t)]σ3 + o

(1

λ

)= e−iνσ3 +O

(1

λ

),

where ν = ν−(x, t) − ν+(x, t) =12

∫ +∞

−∞|q(y, t)|2dy.

On the other hand, substituting the WKB expansion µ±(λ;x, t) =n∑i=0

µ[i]±

(x,t)

λi + O(1λ

)( as λ → ∞) into the

time part of equivalent lax pairs (2.7) and matching the O(λi)(i = 6, 5, 4, 3, 2, 1, 0) in order, combining the O(1) :

µ[0]±,t(x, t) + 4i

[µ[6]± (x, t), σ3

]= (T − T0)µ

[0]± (x, t) and

(µ[0]± (x, t)

)off= 0, one can yield µ

[0]±,t(x, t) = 0, ν±,t(x, t) = 0

and νt = 0 as x → ±∞. That is, ν does not depend on the variable t. Furthermore, for the Zakharov-Shabatspectral problem of the NLS equation, I1 =

∫∞

−∞|u(x, t)|2dx is conservations of mass (also called power), and one

has I1,t = 0 [48]. Here, one can derive I1 = 2ν =∫ +∞

−∞ |q(x, t)|2dx (here x and y in (2.25) are equivalent) is theconservations of mass for the modified Zakharov-Shabat spectral problem of the TOFKN equation, and we also obtainνt = 0 by using the I1,t = 0. Next we give a simple proof of pure algebra for I1,t = 0.

According to the Eq.(1.1), we have

qt =9

2|q|4qx + 3q2|q|2q∗x + 3i|q|2qxx + 3iq2xq

∗ + 3iqq∗xqx − qxxx,

q∗t =9

2|q|4q∗x + 3(q∗)2|q|2qx − 3i|q|2q∗xx − 3i(q∗x)

2q − 3iq∗qxq∗x − q∗xxx,

from I1 =∫ +∞

−∞|q(x, t)|2dx, we have

I1,t =

∫ +∞

−∞

[q(x, t)q∗(x, t)]tdx

=

∫ +∞

−∞

{5

2[q3(q∗)3]x + 3i[|q|2(qxq∗ − qq∗x)]x − [(qxxq

∗ + qq∗xx)x − (qxq∗x)x]

}dx

=5

2q3(q∗)3

∣∣∣+∞

−∞+ 3i|q|2(qxq∗ − qq∗x)

∣∣∣+∞

−∞− (qxxq

∗ + qq∗xx)∣∣∣+∞

−∞+ qxq

∗x

∣∣∣+∞

−∞,

Since q(x, t) = 0 as x→ ±∞, one can derive I1,t = 0 and νt = 0. Therefore, ν is a constant. Proof complete.�


2.1.5. Discrete Spectrum with Double Zeros.

The discrete spectrum of the studied scattering problem is the set of all values λ ∈ C\Σ which satisfy the eigen-functions exist in L2(R). As was proposed by Biondini and Kovacic [47], there are exactly the values of λ in D+

such that s11(λ) = 0 and those values in D− such that s22(λ) = 0. Differing from the previous results with simplepoles [29,34,35], we here suppose that s11(λ) has N double zeros in Λ0 = {λ ∈ C : Reλ > 0, Imλ > 0} denoted by λn,n = 1, 2, · · · , N , that is, s11(λn) = s′11(λn) = 0, and s′′11 6= 0. It follows from the symmetry reduce of the scatteringmatrix that

{s11(±λn) = s22(±λ∗n) = 0,

s′11(±λn) = s′22(±λ∗n) = 0.(2.26)

Therefore, the corresponding discrete spectrum is defined by the set

Λ = {±λn,±λ∗n}Nn=1, (2.27)

whose distributions are shown in Fig. 1. When a given λ0 ∈ Λ∩D+, one can obtain that ψ+1(λ0;x, t) and ψ−2(λ0;x, t)are linearly dependent by combining Eq. (2.11) and s11(λ0) = 0. Similarly, when a given λ0 ∈ Λ∩D−, one can obtainthat ψ+2(λ0;x, t) and ψ−1(λ0;x, t) are linearly dependent by combining Eq. (2.11) and s22(λ0) = 0. For convenience,we introduce the norming constant b[λ0] such that

ψ+1(λ0;x, t) = b[λ0]ψ−2(λ0;x, t), as λ0 ∈ Λ ∩D+,

ψ+2(λ0;x, t) = b[λ0]ψ−1(λ0;x, t), as λ0 ∈ Λ ∩D−.(2.28)

For a given λ0 ∈ Λ∩D+, according to s11(λ0) = det(ψ+1(λ0;x, t), ψ−2(λ0;x, t)) in Eq. (2.11), and taking derivativerespect to λ (here λ = λ0) on both sides of previous equation, we have

s′11(λ0) = det(ψ′+1(λ0;x, t), ψ−2(λ0;x, t)) + det(ψ+1(λ0;x, t), ψ

′−2(λ0;x, t))

= det(ψ′+1(λ0;x, t), ψ−2(λ0;x, t)) + det(b[λ0]ψ−2(λ0;x, t), ψ

′−2(λ0;x, t))

= det(ψ′+1(λ0;x, t), ψ−2(λ0;x, t))− det(b[λ0]ψ

′−2(λ0;x, t), ψ−2(λ0;x, t))

= det(ψ′+1(λ0;x, t)− b[λ0]ψ

′−2(λ0;x, t), ψ−2(λ0;x, t)),

note that s′11(λ0) = 0, ψ′+1(λ0;x, t) − b[λ0]ψ

′−2(λ0;x, t) and ψ−2(λ0;x, t) are linearly dependent. Similarly, for a

given λ0 ∈ Λ ∩D−, one can obtain that ψ′+2(λ0;x, t) − b[λ0]ψ

′−1(λ0;x, t) and ψ−1(λ0;x, t) are linearly dependent by

combining Eq. (2.11) and s′22(λ0) = 0. For convenience, we define another norming constant d[λ0] such that

ψ′+1(λ0;x, t)− b[λ0]ψ

′−2(λ0;x, t) = d[λ0]ψ−2(λ0;x, t), as λ0 ∈ Λ ∩D+,

ψ′+2(λ0;x, t)− b[λ0]ψ

′−1(λ0;x, t) = d[λ0]ψ−1(λ0;x, t), as λ0 ∈ Λ ∩D−.

(2.29)

On the other hand, we notice that ψ+1(λ;x, t) and s11(λ) are analytic on D+, Suppose λ0 is the double zeros ofs11. Let ψ+1(λ;x, t) and s11(λ) carry out Taylor expansion at λ = λ0, we have

ψ+1(λ;x, t)

s11(λ)=

ψ+1(λ0;x, t) + ψ′+1(λ0;x, t)(λ − λ0) +

12!ψ

′′+1(λ0;x, t)(λ− λ0)

2 + · · ·s11(λ0) + s′11(λ0)(λ − λ0) +

12!s

′′11(λ0)(λ − λ0)2 +

13!s

′′′11(λ0)(λ− λ0)3 + · · ·

=ψ+1(λ0;x, t) + ψ′

+1(λ0;x, t)(λ− λ0) +12!ψ

′′+1(λ0;x, t)(λ − λ0)

2 + · · ·12!s

′′11(λ0)(λ− λ0)2 +

13!s

′′′11(λ0)(λ− λ0)3 + · · ·

=2ψ+1(λ0;x, t)

s′′11(λ0)(λ− λ0)

−2 +

(2ψ′

+1(λ0;x, t)

s′′11(λ0)− 2ψ+1(λ0;x, t)s

′′′11(λ0)

3s′′211(λ0)

)(λ− λ0)

−1 + · · · ,

Then, one has the compact form

P−2λ=λ0

[ψ+1(λ;x, t)

s11(λ)

]=

2ψ+1(λ0;x, t)

s′′11(λ0)=

2b[λ0]ψ−2(λ0;x, t)

s′′11(λ0), as λ0 ∈ Λ ∩D+,


where P−2λ=λ0

[f(λ;x, t)] denotes the coefficient of O((λ − λ0)−2) term in the Laurent series expansion of f(λ;x, t) at

λ = λ0. and

Resλ=λ0

[ψ+1(λ;x, t)

s11(λ)

]=

2ψ′+1(λ0;x, t)

s′′11(λ0)− 2ψ+1(λ0;x, t)s

′′′11(λ0)

3s′′211(λ0)=

2(b[λ0]ψ′−2(λ0;x, t) + d[λ0]ψ−2(λ0;x, t))

s′′11(λ0)

− 2b[λ0]ψ−2(λ0;x, t)s′′′11(λ0)

3s′′211(λ0)=

2b[λ0]ψ′−2(λ0;x, t)

s′′11(λ0)+

2d[λ0]ψ−2(λ0;x, t)

s′′11(λ0)− 2b[λ0]ψ−2(λ0;x, t)s

′′′11(λ0)

3s′′211(λ0)

=2b[λ0]ψ

′−2(λ0;x, t)

s′′11(λ0)+

2b[λ0]

s′′11(λ0)· d[λ0]b[λ0]

ψ−2(λ0;x, t)−2b[λ0]

s′′11(λ0)· s

′′′11(λ0)

3s′′11(λ0)ψ−2(λ0;x, t)

=2b[λ0]ψ

′−2(λ0;x, t)

s′′11(λ0)+

[2b[λ0]

s′′11(λ0)

(d[λ0]

b[λ0]− s′′′11(λ0)

3s′′11(λ0)

)]ψ−2(λ0;x, t), as λ0 ∈ Λ ∩D+,

where Resλ=λ0

[f(λ;x, t)] denotes the coefficient of O((λ − λ0)−1) term in the Laurent series expansion of f(λ;x, t) at

λ = λ0.Similarly, for the case of ψ+2(λ;x, t) and s22(λ) are analytic on D−, we repeat the above process and obtain

P−2λ=λ0

[ψ+2(λ;x, t)

s22(λ)

]=

2ψ+2(λ0;x, t)

s′′22(λ0)=

2b[λ0]ψ−1(λ0;x, t)

s′′22(λ0), as λ0 ∈ Λ ∩D−,

Resλ=λ0

[ψ+2(λ;x, t)

s22(λ)

]=

2b[λ0]ψ′−1(λ0;x, t)

s′′22(λ0)+

[2b[λ0]

s′′22(λ0)

(d[λ0]

b[λ0]− s′′′22(λ0)

3s′′22(λ0)

)]ψ−1(λ0;x, t), as λ0 ∈ Λ ∩D−.

Moreover, let

A[λ0] =

2b[λ0]

s′′11(λ0), as λ0 ∈ Λ ∩D+,

2b[λ0]

s′′22(λ0), as λ0 ∈ Λ ∩D−,

B[λ0] =

d[λ0]

b[λ0]− s′′′11(λ0)

3s′′11(λ0), as λ0 ∈ Λ ∩D+,

d[λ0]

b[λ0]− s′′′22(λ0)

3s′′22(λ0), as λ0 ∈ Λ ∩D−.

(2.30)

Then, we have

P−2λ=λ0

[ψ+1(λ;x, t)

s11(λ)

]= A[λ0]ψ−2(λ0;x, t), as λ0 ∈ Λ ∩D+,

P−2λ=λ0

[ψ+2(λ;x, t)

s22(λ)

]= A[λ0]ψ−1(λ0;x, t), as λ0 ∈ Λ ∩D−,

Resλ=λ0

[ψ+1(λ;x, t)

s11(λ)

]= A[λ0][ψ

′−2(λ0;x, t) +B[λ0]ψ−2(λ0;x, t)], as λ0 ∈ Λ ∩D+,

Resλ=λ0

[ψ+2(λ;x, t)

s22(λ)

]= A[λ0][ψ

′−1(λ0;x, t) +B[λ0]ψ−1(λ0;x, t)], as λ0 ∈ Λ ∩D−.

(2.31)

Proposition 7. For λ0 ∈ Λ, the two symmetry relations for A[λ0] and B[λ0] can be deduced as follows:• The first symmetry relation A[λ0] = −A[λ∗0]∗, B[λ0] = B[λ∗0]

∗.• The second symmetry relation A[λ0] = A[−λ∗0]∗, B[λ0] = −B[−λ∗0]∗.

Proof. Considering the first sysmetry relation, for a given λ0 ∈ Λ, s11(λ0) = s22(λ0) = 0, hence substituting S(λ0) =(0 s12(λ0)

s21(λ0) 0

)into ψ+(λ0;x, t) = ψ−(λ0;x, t)S(λ0), one can obtain

ψ+1(λ0;x, t) = s21(λ0)ψ−2(λ0;x, t), as λ0 ∈ Λ ∩D+,

ψ+2(λ0;x, t) = s12(λ0)ψ−1(λ0;x, t), as λ0 ∈ Λ ∩D−.

Therefore, when λ0 ∈ Λ ∩ D+ (λ0 ∈ Λ ∩ D−), one has obtain b[λ0] = s21(λ0) (b[λ0] = s12(λ0)). According to

Eq. (2.18), we have b[λ∗0]∗ = s12(λ

∗0)

∗ = −s21(λ0) = −b[λ0] (where λ∗0 ∈ Λ ∩D−), and A[λ0] =2b[λ0]s′′11(λ0)

=−2b[λ∗

0 ]∗

s′′22(λ∗0)

∗ =

−A[λ∗0]∗. In addition, for a given λ0 ∈ Λ ∩D+ (λ0 ∈ Λ ∩D−), by taking derivative respect to λ (here λ = λ0) bothsides of ψ+1(λ0;x, t) = b[λ0]ψ−2(λ0;x, t) (ψ+2(λ0;x, t) = b[λ0]ψ−1(λ0;x, t)), we have

ψ′+1(λ0;x, t)− b[λ0]ψ

′−2(λ0;x, t) = b′[λ0]ψ−2(λ0;x, t), as λ0 ∈ Λ ∩D+,

ψ′+2(λ0;x, t)− b[λ0]ψ

′−1(λ0;x, t) = b′[λ0]ψ−1(λ0;x, t), as λ0 ∈ Λ ∩D−.


Combining with Eq. (2.29), one can obtain

b[λ0] = s21(λ0) ⇒ d[λ0] = b′[λ0] = s′21(λ0), as λ0 ∈ Λ ∩D+,

b[λ0] = s12(λ0) ⇒ d[λ0] = b′[λ0] = s′12(λ0), as λ0 ∈ Λ ∩D−,

then

B[λ0] =d[λ0]

b[λ0]− s′′′11(λ0)

3s′′11(λ0)=s′21(λ0)

s21(λ0)− s′′′22(λ

∗0)

∗

3s′′22(λ∗0)

∗=

−s′12(λ∗0)∗−s12(λ∗0)∗

− s′′′22(λ∗0)

∗

3s′′22(λ∗0)

∗=

−d[λ∗0]∗−b[λ∗0]∗

− s′′′22(λ∗0)

∗

3s′′22(λ∗0)

∗= B[λ∗0]

∗.

Similarly, by repeating the above process, we can prove the second symmetry relation.�

2.2. Inverse Problem with ZBCs and Double Poles.

In the following parts, we will propose an inverse problem with ZBCs and solve it to obtain accurate double polessolutions for the TOFKN equation (1.1).

2.2.1. The Matrix Riemann-Hilbert Problem.

In general, according to the relation (2.10) of two Jost solutions ψ±(λ;x, t), we will investigate the inverse problemwith ZBCs by establishing a RH problem. In order to pose and solve the RH problem conveniently, we define

ζn =

{λn, n = 1, 2, · · · , N,

−λn−N , n = N + 1, N + 2, · · · , 2N. (2.32)

Then, a matrix RH problem is proposed as follows.

Proposition 8. Define the sectionally meromorphic matrices

M(λ;x, t) =

M+(λ;x, t) =

(µ+1(λ;x, t)

s11(λ), µ−2(λ;x, t)

), as λ ∈ D+,

M−(λ;x, t) =

(µ−1(λ;x, t),

µ+2(λ;x, t)

s11(λ)

), as λ ∈ D−,

(2.33)

where limλ′→λ

λ′∈D±

M(λ′;x, t) =M±(λ;x, t). Then, the multiplicative matrix Riemann-Hilbert problem is given below:

• Analyticity: M(λ;x, t) is analytic in D+∪D−\Λ and has the double poles in Λ, whose principal parts of the Laurentseries at each double pole ζn or ζ∗n, are determined as

Resλ=ζn

M(λ;x, t) =(A[ζn]e

−2iθ(ζn;x,t){µ′−2(ζn;x, t) + [B[ζn]− 2iθ′(ζn;x, t)]µ−2(ζn;x, t)}, 0

),

P−2λ=ζn


−2iθ(ζn;x,t)µ−2(ζn;x, t), 0),

Resλ=ζ∗n

M(λ;x, t) =(0, A[ζ∗n]e

2iθ(ζ∗n;x,t){µ′−1(ζ

∗n;x, t) + [B[ζ∗n] + 2iθ′(ζ∗n;x, t)]µ−1(ζ

∗n;x, t)}

),

P−2λ=ζ∗n

M(λ;x, t) =(0, A[ζ∗n]e

2iθ(ζ∗n;x,t)µ−1(ζ∗n;x, t)

).

(2.34)

• Jump condition:

M−(λ;x, t) =M+(λ;x, t)[I − J(λ;x, t)], as λ ∈ Σ, (2.35)

where

J(λ;x, t) = eiθ(λ;x,t)σ3

(0 −ρ(λ)

ρ(λ) ρ(λ)ρ(λ)

). (2.36)

• Asymptotic behavior:

M(λ;x, t) = eiν−(x,t)σ3 +O

(1

λ

), as λ→ ∞. (2.37)


Proof. For the analyticity of M(λ;x, t), It follows from Eqs. (2.5) and (2.31) that for each double poles ζn ∈ D+ orζ∗n ∈ D−. Now, we consider ζn ∈ D+, and obtain

Resλ=ζn

[ψ+1(ζn;x, t)

s11(ζn)

]= Res

λ=ζn

[µ+1(ζn;x, t)

s11(ζn)eiθ(ζn;x,t)

]= A[ζn][ψ

′−2(ζn;x, t) +B[ζn]ψ−2(ζn;x, t)]

= A[ζn]{[µ−2(ζn;x, t)e−iθ(ζn;x,t)]′ +B[ζn]µ−2(ζn;x, t)e

−iθ(ζn;x,t)},⇒

Resλ=ζn

[µ+1(ζn;x, t)

s11(ζn)

]= A[ζn]e

−2iθ(ζn;x,t){µ′−2(ζn;x, t) + [B[ζn]− 2iθ′(ζn;x, t)]µ−2(ζn;x, t)},⇒

Resλ=ζn

M(λ;x, t) =

(A[ζn]e


),

where the “′” denotes the partial derivative with respect to λ (here λ = ζn), and

P−2λ=ζn

[ψ+1(ζn;x, t)

s11(ζn)

]= P−2λ=ζn

[µ+1(ζn;x, t)


]= A[ζn]µ−2(ζn;x, t)e

−iθ(ζn;x,t),⇒

P−2λ=ζn

[µ+1(ζn;x, t)

s11(ζn)

]= A[ζn]e

−2iθ(ζn;x,t)µ−2(ζn;x, t),⇒

P−2λ=ζn

M(λ;x, t) =

(A[ζn]e

−2iθ(ζn;x,t)µ−2(ζn;x, t), 0

).

Similarly, we also can obtain the analyticity for ζ∗n ∈ D−. It follows from Eqs. (2.5) and (2.10) that{µ+1(λ;x, t) = µ−1(λ;x, t)s11(λ) + µ−2(λ;x, t)e

−2iθ(λ;x,t)s21(λ),

µ+2(λ;x, t) = µ−1(λ;x, t)e2iθ(λ;x,t)s12(λ) + µ−2(λ;x, t)s22(λ),

by combining Eqs. (2.12) and (2.33), one can obtain

M+(λ;x, t) =

(µ+1(λ;x, t)

s11(λ), µ−2(λ;x, t)

)=

(µ−1(λ;x, t) + µ−2(λ;x, t)e

−2iθ(λ;x,t)ρ(λ), µ−2(λ;x, t)

),

M−(λ;x, t) =

(µ−1(λ;x, t),

µ+2(λ;x, t)

s22(λ)

)=

(µ−1(λ;x, t), µ−1(λ;x, t)e

2iθ(λ;x,t)ρ(λ) + µ−2(λ;x, t)

),

and(µ−1(λ;x, t),

µ+2(λ;x, t)

s22(λ)

)=

(µ+1(λ;x, t)

s11(λ), µ−2(λ;x, t)

)(1 e2iθ(λ;x,t)ρ(λ)

−e−2iθ(λ;x,t)ρ(λ) 1− ρ(λ)ρ(λ)

),

M−(λ;x, t) =M+(λ;x, t)(I − J(λ;x, t)),

where J(λ;x, t) is given by Eq. (2.36). The asymptotic behaviors of the modified Jost solutions µ±(λ;x, t) andscattering matrix S(λ) given in Propositions (5) and (6) can easily lead to that of M(λ;x, t). Specifically,

M+(λ;x, t) =

(µ+1(λ;x, t)

s11(λ), µ−2(λ;x, t)

)=

(eiν+(x,t)

e−iν 0

0 e−iν−(x,t)

)+O

(1

λ

)= eiν−(x,t)σ3 +O

(1

λ

), as λ ∈ D+,

M−(λ;x, t) =

(µ−1(λ;x, t),

µ+2(λ;x, t)

s22(λ)

)=

(eiν−(x,t) 0

0 e−iν+(x,t)

eiν

)+O

(1

λ


(1

λ

), as λ ∈ D−.

Therefore, we obtain the asymptotic behavior (2.37) of M(λ;x, t) when λ→ ∞. This completes the proof.�

By subtracting out the asymptotic values as λ → ∞ and the singularity contributions, one can regularize the RHproblem as a normative form. Then, applying the Plemelj’s formula, the solutions of the corresponding matrix RHproblem can be established by an integral equation.

Proposition 9. The solution of the above-mentioned matrix Riemann-Hilbert problem can be expressed as

M(λ;x, t) =eiν−(x,t)σ3 +1

2πi

∫

Σ

M+(ξ;x, t)J(ξ;x, t)

ξ − λdξ+

2N∑

n=1

(Cn(λ)

[µ′−2(ζn;x, t) +

(Dn +

1

λ− ζn

)µ−2(ζn;x, t)

],

Cn(λ)

[µ′−1(ζ

∗n;x, t) +

(Dn +

1

λ− ζ∗n

)µ−1(ζ

∗n;x, t)

]),

(2.38)


where λ ∈ C\Σ,∫Σis an integral along the oriented contour exhibited in Fig. 1, and

Cn(λ) =A[ζn]

λ− ζne−2iθ(ζn;x,t), Cn(λ) =

A[ζ∗n]

λ− ζ∗ne2iθ(ζ

∗n;x,t),

Dn = B[ζn]− 2iθ′(ζn;x, t), Dn = B[ζ∗n] + 2iθ′(ζ∗n;x, t),

(2.39)

µ−k and µ′−k (k = 1, 2) satisfy

µ−1(ζ∗n;x, t) = eiν−(x,t)σ3

(10

)+

2N∑

s=1

Cs(ζ∗n)

[µ′−2(ζs;x, t) +

(Ds +

1

ζ∗n − ζs

)µ−2(ζs;x, t)

]+

1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))1ξ − ζ∗n

dξ,

µ−2(ζs;x, t) = eiν−(x,t)σ3

(01

)+

2N∑

j=1

Cj(ζs)

[µ′−1(ζ

∗j ;x, t) +

(Dj +

1

ζs − ζ∗j

)µ−1(ζ

∗j ;x, t)

]+

1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))2ξ − ζs

dξ,

µ′−1(ζ

∗n;x, t) = −

2N∑

s=1

Cs(ζ∗n)

ζ∗n − ζs

[µ′−2(ζs;x, t) +

(Ds +

2

ζ∗n − ζs

)µ−2(ζs;x, t)

]+

1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))1(ξ − ζ∗n)

2dξ,

µ′−2(ζs;x, t) = −

2N∑

s=1

Cj(ζs)

ζs − ζ∗j

[µ′−1(ζ

∗j ;x, t) +

(Dj +

2

ζs − ζ∗j

)µ−1(ζ

∗j ;x, t)

]+

1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))2(ξ − ζs)2

dξ,

(2.40)

where (M+(ξ;x, t)J(ξ;x, t))j (j = 1, 2) represent the jth column of matrix M+(ξ;x, t)J(ξ;x, t).

Proof. In order to regularize the RH problem, one has to subtract out the asymptotic values as λ→ ∞ which exhibitedin Eq. (2.37) and the singularity contributions. Then, the jump condition (2.35) becomes

M−(λ;x, t) − e−iν−(x,t)σ3 −2N∑

n=1

P−2λ=ζn

M(λ;x, t)

(λ− ζn)2+

Resλ=ζn

M(λ;x, t)

λ− ζn+

P−2λ=ζ∗n

M(λ;x, t)

(λ− ζ∗n)2

+

Resλ=ζ∗n

M(λ;x, t)

λ− ζ∗n

=M+(λ;x, t)

− eiν−(x,t)σ3 −2N∑

n=1

P−2λ=ζn

M(λ;x, t)

(λ− ζn)2+

Resλ=ζn

M(λ;x, t)

λ− ζn+

P−2λ=ζ∗n

M(λ;x, t)

(λ− ζ∗n)2

+

Resλ=ζ∗n

M(λ;x, t)

λ− ζ∗n

−M+(λ;x, t)J(λ;x, t),

(2.41)

where P−2λ=ζn

M(λ;x, t), Resλ=ζn

M(λ;x, t), P−2λ=ζ∗n

M(λ;x, t), Resλ=ζ∗n

M(λ;x, t) have given in Eq. (2.34). By using Plemelj’s

formula, one can obtain the solution (2.38) with formula (2.39) of the matrix RH problem. By combining Eqs. (2.33)and (2.38), µ−1(ζ

∗n;x, t) is the first column element of the solution (2.38) as the double-pole λ = ζ∗n ∈ D−, µ−2(ζs;x, t)

is the second column element of the solution (2.38) as the double-pole λ = ζs ∈ D+. The specific expressions ofµ−1(ζ

∗n;x, t), µ−2(ζs;x, t) and their first derivative to λ are provided in Eq. (2.40).

�

2.2.2. Reconstruction Formula of the Potential.

From the solution (2.38) of the matrix RH problem, we have

M(λ;x, t) = eiν−(x,t)σ3 +M [1](x, t)

λ+O

(1

λ2

), as λ→ ∞, (2.42)

where

M [1](x, t) =− 1

2πi

∫

Σ

M+(ξ;x, t)J(ξ;x, t)dξ +2N∑

n=1

{A[ζn]e

−2iθ(ζn;x,t)[µ′−2(ζn;x, t) +Dnµ−2(ζn;x, t)

],

A[ζ∗n]e2iθ(ζ∗n;x,t)

[µ′−1(ζ

∗n;x, t) + Dnµ−1(ζ

∗n;x, t)

]}.

(2.43)

Substituting Eq. (2.42) into Eq. (2.6) and matching O(λ) term, we have

O(λ) : i[M [1](x, t)σ3 − σ3M

[1](x, t)]= Q(x, t)eiν−(x,t)σ3 ,


then, by expanding the above equation, one can find the reconstruction formula of the double-pole solution (potential)for the TOFKN equation (1.1) with ZBCs as follows

q(x, t) = −2ieiν−(x,t)M[1]12 (x, t), (2.44)

where M[1]12 (x, t) represents the first row and second column element of the matrix M [1](x, t), and

M[1]12 (x, t) = − 1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t)

)12dξ +

2N∑

n=1

{A[ζ∗n]e

2iθ(ζ∗n;x,t)[µ′−11(ζ

∗n;x, t) + Dnµ−11(ζ

∗n;x, t)

]}, (2.45)

where µ−11(ζ∗n;x, t), µ

′−11(ζ

∗n;x, t) represents the first row element of the column vector µ−1(ζ

∗n;x, t), µ

′−1(ζ

∗n;x, t)

respectively. When taking row vector α =(α(1), α(2)

)and column vector γ = (γ(1), γ(2))T, where

α(1) =(A[ζ∗n]e

2iθ(ζ∗n;x,t)Dn

)

1×2N, α(2) =

(A[ζ∗n]e

2iθ(ζ∗n;x,t))

1×2N,

γ(1) =(µ−11(ζ

∗n;x, t)

)1×2N

, γ(2) =(µ′−11(ζ

∗n;x, t)

)1×2N

,(2.46)

we can obtain a more concise reconstruction formulation of the double poles solution (potential) for the TOFKNequation (1.1) with ZBCs and as follows

q(x, t) = −2ieiν−(x,t)

(αγ − 1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t)

)12dξ

). (2.47)

2.2.3. Trace Formulae.

The so-called trace formulae are the scattering coefficients s11(λ) and s22(λ) are formulated in terms of the discretespectrum Λ and reflection coefficients ρ(λ) and ρ(λ). We know that s11(λ), s22(λ) are analytic on D

+, D−, respectively.The discrete spectral points ζn’s are the double zeros of s11(λ), while ζ

∗n’s are the double zeros of s22(λ). Define the

functions β±(λ) as follows:

β+(λ) = s11(λ)

2N∏

n=1

(λ− ζ∗nλ− ζn

)2

eiν ,

β−(λ) = s22(λ)

2N∏

n=1

(λ− ζnλ− ζ∗n

)2

e−iν .

(2.48)

Then, β+(λ) and β−(λ) are analytic and have no zero in D+ and D−, respectively. Furthermore, we have therelation β+(λ)β−(λ) = s11(λ)s22(λ) and the asymptotic behaviors β±(λ) → 1, as λ→ ∞.

According to det(S(λ)) = s11s22 − s21s12 = 1, we can derive

1

s11s22= 1− s21s12

s11s22= 1− ρ(λ)ρ(λ),

by taking logarithm on both sides of the above equation at the same time, that is

−log(s11s22) = log[1− ρ(λ)ρ(λ)] ⇒ log[β+(λ)β−(λ)] = −log[1− ρ(λ)ρ(λ)],

and employing the Plemelj’ formula such that we have

logβ±(λ) = ∓ 1

2πi

∫

Σ

log[1− ρ(λ)ρ(λ)]

ξ − λdξ, λ ∈ D±. (2.49)

Then, substituting Eq. (2.49) into Eq. (2.48), we can obtain the trace formulae

s11(λ) = exp

(− 1

2πi

∫

Σ


ξ − λdξ

)2N∏

n=1

(λ− ζnλ− ζ∗n

)2

e−iν ,

s22(λ) = exp

(1

2πi

∫

Σ


ξ − λdξ

)2N∏

n=1

(λ− ζ∗nλ− ζn

)2

eiν .

(2.50)

2.2.4. Reflectionless Potential: Double-Pole Solitons.

Now, we consider a special kind of reflectionless potential q(x, t) with the reflection coefficients ρ(λ) = ρ(λ) = 0.From the Volterra integral equation (2.9), one obtains ψ±(0;x, t) = µ±(0;x, t) = I, and s11(0) = 1. Combining the

trace formula, one obtains that there exists an integer i ∈ Z such that ν = 8∑Nn=1 arg(ζn) + 2πi.


Then Eqs. (2.40) and (2.47) with J(λ;x, t) = 02×2 become

µ−1(ζ∗n;x, t) = eiν−(x,t)σ3

(10

)+

2N∑

s=1

Cs(ζ∗n)

[µ′−2(ζs;x, t) +

(Ds +

1

ζ∗n − ζs

)µ−2(ζs;x, t)

],

µ−2(ζs;x, t) = eiν−(x,t)σ3

(01

)+

2N∑

j=1

Cj(ζs)

[µ′−1(ζ

∗j ;x, t) +

(Dj +

1

ζs − ζj

)µ−1(ζ

∗j ;x, t)

],

µ′−1(ζ

∗n;x, t) = −

2N∑

s=1

Cs(ζ∗n)

ζ∗n − ζs

[µ′−2(ζs;x, t) +

(Ds +

2

ζ∗n − ζs

)µ−2(ζs;x, t)

],

µ′−2(ζs;x, t) = −

2N∑

s=1

Cj(ζs)

ζs − ζ∗j

[µ′−1(ζ

∗j ;x, t) +

(Dj +

2

ζs − ζ∗j

)µ−1(ζ

∗j ;x, t)

],

(2.51)

and

q(x, t) = −2ieiν−(x,t)αγ. (2.52)

Theorem 10. The explicit expression for the double-pole solution of the TOFKN equation (1.1) with ZBCs is givenby determinant formula

q(x, t) = 2idet(I −G)

detR

(detR

det(I − G)

)2

, (2.53)

where

R =

(0 ατ I −G

), R =

(0 α

τ I − G

),

τ =

(τ (1)

τ (2)

), τ (1) = (1)2N×1, τ (2) = (0)2N×1,

(2.54)

and the 4N × 4N partitioned matrix G =

(G(1,1) G(1,2)

G(2,1) G(2,2)

)with G(i,j) =

(g(i,j)n,j

)

2N×2N(i, j = 1, 2) is given by

g(1,1)n,j =

2N∑

s=1

Cs(ζ∗n)Cj(ζs)

[− 1

ζs − ζ∗j

(Dj +

2

ζs − ζ∗j

)+

(Ds +

1

ζ∗n − ζs

)(Dj +

1

ζs − ζ∗j

)],

g(1,2)n,j =

2N∑

s=1

Cs(ζ∗n)Cj(ζs)

[− 1

ζs − ζ∗j+

(Ds +

1

ζ∗n − ζs

)],

g(2,1)n,j =

2N∑

s=1

Cs(ζ∗n)Cj(ζs)

ζ∗n − ζs

[1

ζs − ζ∗j

(Dj +

2

ζs − ζ∗j

)−(Ds +

2

ζ∗n − ζs

)(Dj +

1

ζs − ζ∗j

)],

g(2,2)n,j =

2N∑

s=1

Cs(ζ∗n)Cj(ζs)

ζ∗n − ζs

[1

ζs − ζ∗j−(Ds +

2

ζ∗n − ζs

)],

and the 4N × 4N partitioned matrix G =

(G(1,1) G(1,2)

G(2,1) G(2,2)

)with G(i,j) =

(g(i,j)n,j

)

2N×2N(i, j = 1, 2) is given by

g(1,1)n,j = ζ∗n

2N∑

s=1

Cs(ζ∗n)Cj(ζs)

ζs

[− 1

ζs − ζ∗j

(Dj +

2

ζs − ζ∗j

)+ζsζ∗j

(Ds +

1

ζ∗n − ζs− 1

ζs

)(Dj +

1

ζs − ζ∗j− 1

ζ∗j

)],

g(1,2)n,j = ζ∗n

2N∑

s=1

Cs(ζ∗n)Cj(ζs)

ζs

[− 1

ζs − ζ∗j+ζsζ∗j

(Ds +

1

ζ∗n − ζs− 1

ζs

)],

g(2,1)n,j =

2N∑

s=1

Cs(ζ∗n)Cj(ζs)

ζ∗n − ζs

[1

ζs − ζ∗j

(Dj +

2

ζs − ζ∗j

)− ζsζ∗j

(Ds +

2

ζ∗n − ζs

)(Dj +

1

ζs − ζ∗j− 1

ζ∗j

)],

g(2,2)n,j =

2N∑

s=1

Cs(ζ∗n)Cj(ζs)

ζ∗n − ζs

[1

ζs − ζ∗j− ζsζ∗j

(Ds +

2

ζ∗n − ζs

)].

Proof. From the Eqs. (2.46), (2.51) and (2.52), the reflectionless potential is derived by determinant formula:

q(x, t) = 2idet(R)

det(I −G)e2iν−(x,t). (2.55)


(a)-6 -4 -2 0 2 4 6

x

-15

-10

-5

0

5

10

15

t

(b) (c)

Figure 2. (Color online) The 1-double-pole soliton solution of TOFKN equation (1.1) with ZBCsand N = 1, ζ1 = 1

2 + 12 i, A[ζ1] = 1, B[ζ1] = 1. (a) The three-dimensional plot; (b) The density plot;

(c) The sectional drawings at t = −6 (dashed line), t = 0 (solid line), and t = 6 (dash-dot line).

However, this formula (2.55) is implicit due to unknown function ν−(x, t) is included. We need to deduce an exactform for the reflectionless potential. From the trace formulae (2.50) and Volterra integral equation (2.9) as x→ −∞,we get

M(λ;x, t) = I + λ

2N∑

n=1

P−2λ=ζn

(M(λ;x, t)/λ

)

(λ− ζn)2+

Resλ=ζn

(M(λ;x, t)/λ

)

λ− ζn+

P−2λ=ζ∗n

(M(λ;x, t)/λ

)

(λ− ζ∗n)2

+

Resλ=ζ∗n

(M(λ;x, t)/λ

)

λ− ζ∗n

,

(2.56)

which can yield the γ given by Eq. (2.46) explicitly. Then, substituting γ into the formula of the potential, one yields

q(x, t) = 2idet(R)

det(I − G)eiν−(x,t), (2.57)

then, by combining Eq. (2.56) with Eq. (2.57), we can obtain the determinant formula (2.53). Complete the proof.�

For example, we obtain the N -double-pole solutions of the TOFKN equation (1.1) with ZBCs via Eq. (2.53):• When taking parametersN = 1, ζ1 = 1

2+12 i, A[ζ1] = 1, B[ζ1] = 1, we can obtain the explicit 1-double-pole solution

and give out relevant plots in Fig. 2. Figs. 2 (a) and (b) exhibit the three-dimensional and density diagrams for theexact 1-double-pole soliton solution of the TOFKN equation with ZBCs, which is equivalent to the elastic collisionsof two bright-bright solitons. Fig, 2 (c) displays the distinct profiles of the exact 1-double-pole soliton solution att = ±6, 0.

Compared with the classical second-order flow KN system which also be called the DNLS equation in Ref. [44], thecomparison made between the density diagrams of the 1-double-pole soliton solutions for the TOFKN equation andthe DNLS equation shows that the trajectories of solutions are different obviously, it also means the introduction ofthird-order dispersion and quintic nonlinear term of KN systems can affect the trajectories of solutions. By comparingwith Ref. [44], we find that the wave heights of the 1-double-pole solution at x = 0 and t = 0 are consistent, andthe wave heights are all 1168/157. Therefore, we make a guess that the energy at t = 0 of the central peak for the1-double-pole solution of the TOFKN equation is the same with that corresponding to the DNLS equation by selectingthe same parameters. In other words, third-order dispersion and quintic nonlinear term of KN systems have littleeffect on the maximum amplitude of the solutions.

Remark 11. The parameter selection of the 1-double-pole soliton solution shown in Fig. 2 is consistent with that inreference [44].

• When taking parameters N = 2, ζ1 = 12 + 1

2 i, A[ζ1] = B[ζ1] = 1, ζ2 = 13 + 1

2 i, A[ζ2] = B[ζ2] = 1, we can obtainthe explicit 2-double-pole solution and give out relevant plots in Fig. 3. Figs. 3 (a) and (b) exhibit the three-dimensional and density diagrams for the exact 2-double-pole soliton solution of the TOFKN equation with ZBCs,which is equivalent to the interaction of two 1-double-pole soliton solutions. Fig. 3 (c) displays the distinct profiles ofthe exact 2-double-pole soliton solution at t = ±20, 0.

3. The IST with NZBCs and Double Poles Solution


(a)-25 -20 -15 -10 -5 0 5 10 15 20 25

x

-30

-20

-10

0

10

20

30

t

(b) (c)

Figure 3. (Color online) The 2-double-pole soliton solution of TOFKN equation (1.1) with ZBCsand N = 2, ζ1 = 1

2 + 12 i, A[ζ1] = B[ζ1] = 1, ζ2 = 1

3 + 12 i, A[ζ2] = B[ζ2] = 1. (a) The three-dimensional

plot; (b) The density plot; (c) The sectional drawings at t = −20 (dashed line), t = 0 (solid line), andt = 20 (dash-dot line).

In this section, we will find the double-pole solution q(x, t) for the TOFKN equation (1.1) with the NZBCs

q(x, t) ∼ q±, as x→ ±∞, (3.1)

where |q±| = q0 > 0, and q± are independent of x, t.

3.1. Direct Scattering with NZBCs and Double Poles.

In this subsection, we will investigate direct scattering with NZBCs and the case of double poles for the modifiedZakharov-Shabat eigenvalue problem of TOFKN equation (1.1) in detail.

3.1.1. Jost Solutions, Analyticity, and Continuity.

As x → ±∞, we consider the asymptotic scattering problem of the modified Zakharov-Shabat eigenvalue problem(1.2)-(1.3):

Ψx = X±Ψ, Ψt = T±Ψ, X± = iλ2σ3 + λQ±,

T± =

(4λ4 +

3

2q40 − 2λ2q20

)X±, Q± =

(0 q±

−q∗± 0

).

(3.2)

Proposition 12. the fundamental matrix solution of Eq. (3.2) is presented as follows:

Ψbg± (λ;x, t) =

Y±(λ)eiθ(λ;x,t)σ3 , λ 6= ±iq0, and λ+ k(λ) 6= 0,

I +

(x+

15

2q40t

)X±(λ), λ = ±iq0,

(3.3)

where

Y±(λ) =

(1 iq±

ziq∗±z

1

)= I +

i

zσ3Q±, z = λ+ k(λ),

θ(λ;x, t) = λk(λ)

[x+

(4λ4 +

3

2q40 − 2λ2q20

)t

], k2(λ) = λ2 + q20 .

(3.4)

Proof. For the case of λ = ±iq0, we can directly calculate and obtain Ψbg± (λ;x, t) = I +

(x+ 15

2 q40t)X±(λ). However,

when λ 6= ±iq0, based on eigenvalues ±iλk(λ), one can derive k2(λ) = λ2 + q20 , and the eigenvector matrix of X± andT± can be written

Y±(λ) =

(1 iq±

λ+k(λ)iq∗±

λ+k(λ) 1

)=

(1 iq±

ziq∗±z

1

)= I +

i

zσ3Q±, z = λ+ k(λ) 6= 0. (3.5)

Now, X± and T± can are diagonalized by the eigenvector matrix (3.5) as

X± = Y±[iλk(λ)σ3]Y−1± , T± = Y±

[(4λ2 +

3

2q40 − 2λ2q20

)iλk(λ)σ3

]Y −1± , (3.6)

and direct calculation shows that

detY± = 1 +q20z2

△= η, (3.7)


and

Y −1± =

1

η

(1 − iq±

z

− iq∗±z

1

)=

1

η

(I − i

zσ3Q±

). (3.8)

By substituting (3.6) into (3.2), we have

(Y −1± Ψ

)x= iλk(λ)σ3

(Y −1± Ψ

),(Y −1± Ψ

)t=

(4λ2 +

3

2q40 − 2λ2q20

)iλk(λ)σ3

(Y −1± Ψ

), (3.9)

from which we can derive the solution of the asymptotic spectral problem (3.2)

Ψ = Ψbg± (λ;x, t) = Y±(λ)e

iθ(λ;x,t)σ3 , (3.10)

where θ(λ;x, t) is given by (3.4). This completes the proof.�

Since k(λ) stands for a two-sheeted Riemann surface. Therefore, in order to avoid multi-valued case of eigenvaluek(λ), we define a new uniformization variable: z = λ+ k(λ), which was first introduced by Faddeev and Takhtajan in1987 [49]. Next, we will illustrate the scattering problem on a standard z-plane instead of the two-sheeted Riemannsurface by utilizing two single-valued inverse mappings:

k(λ) =1

2

(z +

q20z

), λ =

1

2

(z − q20

z

), (3.11)

where the inverse mapping evidently satisfy the relation k2(λ) = λ2 + q20 . It means that the Jost solutions ψ±(z;x, t)of the Lax pairs (1.2)-(1.3) possess the following asymptotics:

ψ±(z;x, t) ∼ Y±(z)eiθ(z;x,t)σ3, as x→ ±∞, (3.12)

and the modified Jost solutions µ±(z;x, t) are derived by making the transform

µ±(z;x, t) = ψ±(z;x, t)e−iθ(z;x,t)σ3 , (3.13)

such that

µ±(z;x, t) ∼ Y±(z), as x→ ±∞. (3.14)

On the other hand, the modified Jost solutions µ±(z;x, t) satisfy the following equivalent Lax pair:[Y −1± (z)µ±(z;x, t)

]

x+ iλk(λ)

[Y −1± (z)µ±(z;x, t), σ3

]= Y −1

± (z)∆X±µ±(z;x, t), (3.15)

[Y −1± (z)µ±(z;x, t)

]

t+ iλk(λ)

(4λ4 +

3

2q40 − 2λ2q20

)[Y −1± (z)µ±(z;x, t), σ3

]= Y −1

± (z)∆T±µ±(z;x, t), (3.16)

where ∆X± = X(λ(z);x, t

)−X±

(λ(z);x, t

)= λ(z)[Q(x, t) −Q±(x, t)] := λ(z)∆Q±(x, t) and ∆T± = T

(λ(z);x, t

)−

T±(λ(z);x, t

). Equations (3.15) and (3.16) can be written in full derivative form

d(e−iθ(z;x,t)σ3Y −1

± (z)µ±(z;x, t))= e−iθ(z;x,t)σ3

[Y −1± (z)

(∆X±dx+∆T±dt

)µ±(z;x, t)

], (3.17)

which lead to Volterra integral equations

µ±(z;x, t) = Y±(z) +

λ(z)

∫ x

±∞

Y±(z)eik(z)λ(z)(x−y)σ3

[Y −1± (z)∆Q±(y, t)µ±(z; y, t)

]dy, z 6= 0,±iq0,

λ(z)

∫ x

±∞

[I + (x− y)X±(z)

]∆Q±(y, t)µ±(z; y, t)

]dy, z = ±iq0.

(3.18)

Since the exponential function contains k(z)λ(z), we define the z plane by the positive and negative values ofIm(k(z)λ(z)

), that is

k(z)λ(z) =1

4

(z2 − q40

z2

)=

|z|4z2 − q40z2 + q40z

2 − q40(z∗)2

4|z|4 =1

4|z|4[(|z|4 − q40)z

2 + 4iq40RezImz],

Im[k(z)λ(z)

]=

1

4|z|4[(2|z|4 − 2q40)RezImz + 4q40RezImz

]=

1

4|z|4[(2|z|4 + 2q40)RezImz

],

and

Im[k(z)λ(z)

]> 0, that is RezImz > 0,

Im[k(z)λ(z)

]< 0, that is RezImz < 0,

Im[k(z)λ(z)

]= 0, that is z4 = q40 ⇒

{z = ±iq0,z = ±q0.

Therefore, from the above mapping relation between the λ-plane and z-plane, we define Σ and D± on the z-planeas Σ := R ∪ iR\{0}, D± := {z ∈ C| ± (Rez)(Imz) > 0}, which are shown in Fig. 4.


Rez

Imz

0

zn

z∗n

−z∗n

−zn

− q20zn

− q20z∗n

q20z∗n

q20zn

−ω∗m

−ωm

ωm

ω∗m

Figure 4. (Color online) Distribution of the discrete spectrum and jumping curves for the Riemann-Hilbert problem on complex z-plane, Region D+ =

{k ∈ C

∣∣RezImz > 0}

(yellow region), region

D− ={z ∈ C

∣∣RezImz < 0}(white region).

Then, according to the Volterra integral equations (3.18) for the case of z 6= 0,±iq0, the following analyticity of the(modified) Jost solution can be derived as follows:

Proposition 13. Suppose (1+|x|)(q(x, t)−q±

)∈ L1(R±). Then, Jost solution ψ(z;x, t) (modified Jost solution µ(z;x, t))

have the following properties:• Eq. (1.2) has the unique solution ψ(z;x, t) (µ(z;x, t)) satisfying Eq. (3.12) (Eq. (3.14)) on Σ.• The column vectors ψ+1(z;x, t) (µ+1(z;x, t)) and ψ−2(z;x, t) (µ−2(z;x, t)) can be analytically extended to D+ andcontinuously extended to D+ ∪ Σ.• The column vectors ψ+2(z;x, t) (µ+2(z;x, t)) and ψ−1(z;x, t) (µ−1(z;x, t)) can be analytically extended to D− andcontinuously extended to D− ∪ Σ.

Proof. We define modified Jost solutions µ± = (µ±1, µ±2) =

(µ±11 µ±21

µ±12 µ±22

), in which µ±1 and µ±2 is the first and

second columns of µ±, respectively. Then, taking µ− as an example, Eq. (3.18) (here z 6= 0,±iq0) can be rewritten as(µ−11 µ−21

µ−12 µ−22

)=

(1 iq−

ziq∗−z

1

)+ λ

∫ x

−∞

(− 1q20+z

2

[Ξ1e

−2ik(z)λ(z)(x−y) + Ξ2

]1

q20+z2

[Ξ3e

2ik(z)λ(z)(x−y) + Ξ4

]

− 1q20+z

2

[Ξ5e

−2ik(z)λ(z)(x−y) + Ξ6

]1

q20+z2

[Ξ7e

2ik(z)λ(z)(x−y) + Ξ8

])dy,

where

Ξ1 = iq∗µ−11q−z − iµ−11q20z − qµ−12q

20 + µ−12q

20q−,

Ξ2 = −iq∗µ−11q−z + iµ−11q20z − qµ−12z

2 + µ−12q−z2,

Ξ3 = iq∗µ−21q−z − iµ−21q20z + qµ−22z

2 − µ−22q−z2,

Ξ4 = −iq∗µ−21q−z + iµ−21q20z + qµ−22q

20 − µ−22q

20q−,

Ξ5 = iqµ−12q∗−z − iµ−12q

20z + q∗µ−11z

2 − µ−11q∗−z

2,

Ξ6 = −iqµ−12q∗−z + iµ−12q

20z + q∗µ−11q

20 − µ−11q

20q

∗−,

Ξ7 = iqµ−22q∗−z − iµ−22q

20z − q∗µ−21q

20 + µ−21q

20q

∗−,

Ξ8 = −iqµ−22q∗−z + iµ−22q

20z − q∗µ−21z

2 + µ−21q∗−z

2.

Note that µ−1 =

(1iq∗−z

)+ λ(z)

∫ x−∞

(− 1q20+z

2

[Ξ1e

−2ik(z)λ(z)(x−y) + Ξ2

]

− 1q20+z

2

[Ξ5e

−2ik(z)λ(z)(x−y) + Ξ6

])dy, where e−2ik(z)λ(z)(x−y) =

e−2i(x−y)Re[k(z)λ(z)]e2(x−y)Im[k(z)λ(z)]. Since x − y > 0 and Im[k(z)λ(z)

]< 0, that is RezImz < 0 , it indicates that

the first column of µ− is analytically extended to D−. In addition, since Im[k(z)λ(z)] = 0 when z ∈ Σ, it demonstrate

that the first column of µ− is continuously extended to D− ∪ Σ. In the same way, we also obtain the analyticityand continuity of µ−2, µ+1 and µ+2. Similarly, the analyticity and continuity for the Jost solutions ψ±(λ;x, t) can besimply presented via the analyticity and continuity of µ±(λ;x, t) and the relation (3.13). This completes the proof.

�


Similar to Proposition 2 in Sec. 2 with ZBCs, one can also confirm that the Jost solutions ψ(z;x, t) are thesimultaneous solutions for both parts of the modified Zakharov-Shabat eigenvalue problem Eqs. (1.2)-(1.3) withNZBCs.

3.1.2. Scattering Matrix, Analyticity, and Continuity.

Since ψ±(z;x, t) are two fundamental matrix solutions of the modified Zakharov-Shabat eigenvalue problem Eqs.(1.2)-(1.3) in z ∈ Σ\{±iq0}, there exists a linear relation between ψ+(z;x, t) and ψ−(z;x, t), so one can define theconstant scattering matrix S(z)

(sij(z)

)2×2

such that

ψ+(z;x, t) = ψ−(z;x, t)S(z), z ∈ Σ\{±iq0}. (3.19)

According to the (2.11) of the derivation process, and combining with formula det(ψ±(z;x, t)

)= det(Y±) = η, the

scattering coefficients shown as:

s11(z) =det(ψ+1(z;x, t), ψ−2(z;x, t)

)

η, s12(z) =

det(ψ+2(z;x, t), ψ−2(z;x, t)

)

η,

s21(z) =det(ψ−1(z;x, t), ψ+1(z;x, t)

)

η, s22(z) =

det(ψ−1(z;x, t), ψ+2(z;x, t)

)

η.

(3.20)

From these determinant representations, one can extend the analytical regions of s11(z) and s22(z).

Proposition 14. Suppose that q(x, t) − q± ∈ L1(R±). Then, s11(λ) can be analytically extended to D+ and contin-uously extended to D+ ∪

(Σ\{±iq0}

), while s22(λ) can be analytically extended to D− and continuously extended to

D− ∪(Σ\{±iq0}

). Moreover, both s12(λ) and s21(λ) are continuous in Σ\±iq0.

Proof. The proposition can be verified by using Proposition (13) and Eq. (3.20). �

Proposition 15. Suppose that (1 + |x|)(q(x, t)− q±) ∈ L1(R±). Then, k(z)s11(λ) can be analytically extended to D+

and continuously extended to D+ ∪Σ, while k(z)s22(λ) can be analytically extended to D− and continuously extendedto D− ∪Σ. Moreover, both k(z)s12(λ) and k(z)s21(λ) are continuous in Σ.

Proof. The proposition can be verified by using Proposition (13) and Eqs. (3.7), (3.11) and (3.20). �

In order to study the RH problem in the inverse process, we focus on the potential without spectral singularity,and suppose sij(z)(i, j = 1, 2) are continuous in the branch points {±iq0}. The reflection coefficients ρ(λ) and ρ(λ)are defined as follows:

ρ(z) =s21(z)

s11(z), ρ(z) =

s12(z)

s22(z), z ∈ Σ, (3.21)

which will be utilized in the inverse scattering problem.

3.1.3. Symmetry Structures.

Compared with the case of ZBCs, The symmetries of X(z;x, t), T (z;x, t), Jost solutions, modified Jost solutions,scattering matrix and reflection coefficients of the case of NZBCs will be more complicated.

Proposition 16. For the case of NZBCs, the Jost solutions, modified Jost solutions, scattering matrix and reflectioncoefficients admit following three kinds of reduction conditions on the z-plane:• The first symmetry reduction

X(z;x, t) = σ2X(z∗;x, t)∗σ2, T (z;x, t) = σ2T (z∗;x, t)∗σ2,

ψ±(z;x, t) = σ2ψ±(z∗;x, t)∗σ2, µ±(z;x, t) = σ2µ±(z

∗;x, t)∗σ2,

S(z) = σ2S(z∗)∗σ2, ρ(z) = −ρ(z∗)∗.

(3.22)

• The second symmetry reduction

X(z;x, t) = σ1X(−z∗;x, t)∗σ1, T (z;x, t) = σ1T (−z∗;x, t)∗σ1,ψ±(z;x, t) = σ1ψ±(−z∗;x, t)∗σ1, µ±(z;x, t) = σ1µ±(−z∗;x, t)∗σ1,

S(z) = σ1S(−z∗)∗σ1, ρ(z) = ρ(−z∗)∗.(3.23)


• The third symmetry reduction

X(z;x, t) = X

(− q20

z;x, t

), T (z;x, t) = T

(− q20

z;x, t

),

ψ±(z;x, t) =i

zψ±

(− q20

z;x, t

)σ3Q±, µ±(z;x, t) =

i

zµ±

(− q20

z;x, t

)σ3Q±,

S(z) = (σ3Q−)−1S

(− q20

z

)σ3Q+, ρ(z) =

q∗−q−ρ

(− q20

z

).

(3.24)

Proof. The proof of the first symmetry reduction and the second symmetry reduction are similar to Proposition (4).We only need to replace λ in Proposition (4) with z. Next, we mainly prove the third symmetry reduction.

Since λ(z) = 12

(z − q20

z

), one can derive λ

(− q20

z

)= 1

2

(− q20

z+ z

)= λ(z), we have

X(λ(z);x, t) = X

(λ

(− q20

z

);x, t

)⇒ X(z;x, t) = X

(− q20

z;x, t

),

T (λ(z);x, t) = T

(λ

(− q20

z

);x, t

)⇒ T (z;x, t) = T

(− q20

z;x, t

). (3.25)

Then ψ±

(− q20

z;x, t

)C is also the solution of Eqs. (1.2)-(1.3), for any 2 × 2 matrix C independent of x and t.

Hence, according to (3.12), it is apparent that

ψ±

(− q20

z;x, t

)C ∼ Y±

(− q20

z

)eiθ(−

q20z;x,t)σ3C = Y±

(− q20

z

)e−iθ(z;x,t)σ3C, as x→ ±∞, (3.26)

where θ(− q20

z;x, t

)= −θ(z;x, t), and noting that

i

zY±

(− q20

z

)e−iθ(z;x,t)σ3σ3Q± = Y±(z)e

iθ(z;x,t)σ3 ,

which together with (3.26) implies that C = izσ3Q±, we then obtain the symmetric relation

ψ±(z;x, t) =i

zψ±

(− q20

z;x, t

)σ3Q±, (3.27)

and by using the transformation (3.13), we have

µ±(z;x, t) =i

zµ±

(− q20

z;x, t

)eiθ(−

q20z;x,t)σ3σ3Q±e

−iθ(z;x,t)σ3 =i

zµ±

(− q20

z;x, t

)e−iθ(z;x,t)σ3σ3Q±e

−iθ(z;x,t)σ3

=i

zµ±

(− q20

z;x, t

)σ3Q±.

In addition, By using Eq. (3.19), we have

ψ+

(− q20

z;x, t

)= ψ−

(− q20

z;x, t

)S

(− q20

z

). (3.28)

and according the symmetry of Jost solutions (3.27), one has following formulas:

ψ+

(− q20

z;x, t

)=z

iψ+(z;x, t)(σ3Q+)

−1, ψ−

(− q20

z;x, t

)=z

iψ−(z;x, t)(σ3Q−)

−1, (3.29)

then substituting (3.29) into Eq. (3.28), one can obtain

z

iψ+(z;x, t)(σ3Q+)

−1 =z

iψ−(z;x, t)(σ3Q−)

−1S

(− q20

z

)⇒ ψ+(z;x, t) = ψ−(z;x, t)(σ3Q−)

−1S

(− q20

z

)σ3Q+,

(3.30)

combining (3.30) with (3.19), and considering the same asymptotic behavior,

S(z), S

(− q20

z

)∼ I, as x→ ±∞,

one can lead to

S(z) = (σ3Q−)−1S

(− q20

z

)σ3Q+. (3.31)


According to the symmetry reduction (3.31) of scattering matrix, we have

(s11(z) s12(z)s21(z) s22(z)

)=

((1 00 −1

)(0 q−

−q∗− 0

))−1

s11

(− q20

z

)s12

(− q20

z

)

s21

(− q20

z

)s22

(− q20

z

)

(1 00 −1

)(0 q+

−q∗+ 0

)

=

q∗+q∗−

s22

(− q20

z

)q+q∗−

s21

(− q20

z

)

q∗+q−s12

(− q20

z

)q+q−s11

(− q20

z

)

,

we then obtain these symmetric relations

s11(z) =q∗+q∗−s22

(− q20

z

), s12(z) =

q+q∗−s21

(− q20

z

), s21(z) =

q∗+q−s12

(− q20

z

), s22(z) =

q+q−s11

(− q20

z

). (3.32)

Since ρ(z) = s21(z)s11(z)

=

q∗+q−s12

(−

q20z

)

q∗+

q∗−s22

(−

q20z

) =q∗−q−ρ(− q20

z

), the symmetry reduction of reflection coefficient is ρ(z) =

q∗−q−ρ(− q20

z

). This completes the proof.

�

3.1.4. Asymptotic Behaviors.

In order to propose and solve the matrix RH problem for the inverse problem in the next part, it is necessary todiscuss the asymptotic behaviors of the modified Jost solutions and scattering matrix as z → 0 and z → ∞, whichdiffer from the case of ZBCs. The asymptotic behaviors of the modified Jost solution are deduce with the aid of thegeneral WKB expansions.

Proposition 17. The asymptotic behaviors of the modified Jost solutions are exhibited as

µ±(z;x, t) =

i

zeiν±(x,t)σ3σ3Q± +O(1), as z → 0,

eiν±(x,t)σ3 +O

(1

z

), as z → ∞,

(3.33)

where

ν±(x, t) =1

2

∫ x

±∞

(|q(y, t)|2 − q20

)dy. (3.34)

Proof. First, we consider the following asymptotic WKB expansion of the modified Jost solutions µ±(z;x, t) as z → 0

µ±(z;x, t) =

n∑

i=−1

µ[i]± (x, t)zi +O

(zn+1

)=µ[−1]± (x, t)

z+ µ

[0]± (x, t) + µ

[1]± (x, t)z + · · · . (3.35)

Substituting (3.35) into the equivalent Lax pair (3.15), and comparing the coefficients of different powers of z, wehave

O(z−4) :q404

[σ3Q±µ

[−1]± (x, t), σ3

]= 0,⇒

(µ[−1]± (x, t)

)diag= 0,

where(µ[−1]± (x, t)

)diagrepresent the diagonal parts of µ

[−1]± (x, t).

O(z−3) : −q40

4i[µ[−1]± (x, t), σ3

]− q40

4

[σ3Q±µ

[0]± (x, t), σ3

]=q202iσ3Q±(Q −Q±)µ

[−1]± (x, t),

⇒(µ[0]± (x, t)

)diag= diag

(−iqq20

µ[−1]±21(x, t),

−iq∗q20

µ[−1]±12(x, t)

),

where µ[−1]±21(x, t) and µ

[−1]±12(x, t) represent the second row and first column element and the first row and second column

element of matrix µ[−1]± (x, t), respectively. Moreover,

diag

(−iqq20

µ[−1]±21(x, t),

−iq∗q20

µ[−1]±12(x, t)

)=

(−iqq20µ[−1]±21(x, t) 0

0 −iq∗

q20µ[−1]±12(x, t)

).


O(z−2) :(−iσ3Q±µ

[−1]± (x, t)

)

x− q40

4i[µ[0]± (x, t), σ3

]− q40

4

[σ3Q±µ

[−1]± (x, t), σ3

]= −q

20

2(Q−Q±)µ

[−1]± (x, t)+

q202iσ3Q±(Q −Q±)µ

[0]± (x, t),⇒

(µ[−1]± (x, t)

)

x=

1

2i(|q(x, t)|2 − q20

)σ3µ

[−1]± (x, t).

Therefore, we obtain that µ[−1]± (x, t) = e

∫x

±∞i2 (|q(y,t)|

2−q20)dyσ3C, where C is off-diagonal constant matrix. Let

ν±(x, t) satisfy Eq. (3.34), one can obtain µ[−1]± (x, t) = eiν±(x,t)σ3C and lim

x→±∞µ[−1]± (x, t) = C. According to the

expansion (3.35), we have

limx→±∞

zµ±(z;x, t) = zY±(z) = z

(I +

i

zσ3Q±

)= limx→±∞

(µ[−1]± (x, t) + zµ

[0]± (x, t) + · · ·

),

so we have C = iσ3Q± and µ[−1]± (x, t) = ieiν±(x,t)σ3σ3Q±. Finally, we obtain the asymptotic behaviors

µ±(z;x, t) =i

zeiν±(x,t)σ3σ3Q± +O(1), as z → 0.

Next, we consider the asymptotic expansions of the Jost solutions µ±(z;x, t) as z → ∞

µ±(z;x, t) =n∑

i=0

µ[i]± (x, t)

zi+O

(1

zn+1

)= µ

[0]± (x, t) +

µ[1]± (x, t)

z+µ[2]± (x, t)

z2+ · · · . (3.36)

By utilizing same way, substituting (3.36) into the equivalent Lax pair (3.15), and comparing the coefficients ofdifferent powers of z, we have

O(z2) :i

4

[µ[0]± (x, t), σ3

]= 0,⇒

(µ[0]± (x, t)

)off= 0,

where(µ[0]± (x, t)

)offrepresent the off-diagonal parts of µ

[0]± (x, t).

O(z) :i

4

[µ[1]± (x, t), σ3

]+

1

4

[σ3Q±µ

[0]± (x, t), σ3

]=

1

2(Q−Q±)µ

[0]± (x, t),⇒

(µ[1]± (x, t)

)off= off

(iqµ

[0]±22(x, t), iq

∗µ[0]±11(x, t)

),

where off(iqµ

[0]±22(x, t), iq

∗µ[0]±11(x, t)

)=

(0 iqµ

[0]±22(x, t)

iq∗µ[0]±11(x, t) 0

).

O(1) :µ[0]±,x(x, t) +

i

4

[µ[2]± (x, t), σ3

]+

1

4

[σ3Q±µ

[1]± (x, t), σ3

]=

1

2(Q−Q±)µ

[1]± (x, t)− i

2σ3Q±(Q −Q±)µ

[0]± (x, t),

⇒(µ[0]± (x, t)

)

x=i

2

(|q(x, t)|2 − q20

)σ3.

Therefore, we obtain that µ[0]± (x, t) = Ce

∫x

±∞i2 (|q(y,t)|

2−q20)dyσ3 , where C is diagonal constant matrix. Let ν±(x, t)

satisfy Eq. (3.34), one can obtain µ[0]± (x, t) = Ceiν±(x,t)σ3 and lim

x→±∞µ[0]± (x, t) = C. According to the expansion (3.36),

we have

limx→±∞

z→∞

µ±(z;x, t) = limz→∞

Y±(z) = limz→∞

(I +

i

zσ3Q±

)= I,

so we have C = I and µ[0]± (x, t) = eiν±(x,t)σ3 . Finally, we get the asymptotic behaviors

µ±(z;x, t) = eiν±(x,t)σ3 + O

(1

z

), as z → ∞.

This completes the proof. �

The asymptotic behaviors of the scattering matrix can be directly yielded by exploiting the determinant represen-tations of the scattering matrix and the asymptotic behaviors of the modified Jost solutions.

Proposition 18. The asymptotic behaviors of the scattering matrix are as follows

S(z) =

diag

(q−q+,q+q−

)eiν(t)σ3 +O(z), as z → 0,

e−iν(t)σ3 +O

(1

z

), as z → ∞,

(3.37)

where

ν(t) =1

2

∫ +∞

−∞

(|q(y, t)|2 − q20

)dy. (3.38)


Proof. From the relationship (3.13) between Jost solutions and modified Jost solutions, the determinant representations(3.20) of the scattering matrix and the asymptotic behaviors (3.33), for z → 0, we have

s11(z) =det(ψ+1(z;x, t), ψ−2(z;x, t)

)

η=

det(µ+1(z;x, t), µ−2(z;x, t)

)

1 +q20z2

=

det

(O(1) i

zq−e

−iν−(x,t) +O(1)izq∗+e

iν+(x,t) +O(1) O(1)

)(z2

q20− z4

q40+ · · ·

)

=

{q−q

∗+

z2e−i[ν−(x,t)−ν+(x,t)] +O(z)

}(z2

q20− z4

q40+ · · ·

)=q−q+

e−iν(t) +O(z),

where 1

1+q20z2

= z2

z2+q20= z2

q20· 1

1−

(− z2

q20

) = z2

q20− z4

q40+ · · · , and ν(t) = ν−(x, t) − ν+(x, t) = 1

2

∫ +∞

−∞

(|q(y, t)|2 − q20

)dy.

Similarly, one can obtain

s12 = O(z), s21 = O(z), s22 =q+q−

eiν(t) +O(z),

so we have the asymptotic behavior (3.37) as z → 0.When z → ∞, by using the same way, we have

s11(z) =det(ψ+1(z;x, t), ψ−2(z;x, t)

)

η=

det(µ+1(z;x, t), µ−2(z;x, t)

)

1 +q20z2

= det

(eiν+(x,t) +O(z−1) O(z−1)

O(z−1) e−iν−(x,t) +O(z−1)

)

(1−

q20z2

+q40z4

+ · · ·

)=

{e−i[ν−(x,t)−ν+(x,t)] +O(z−1)

}(1−

q20z2

+q40z4

+ · · ·

)= e−iν(t) +O(z−1),

where 1

1+q20z2

= 1

1−

(−

q20z2

) = 1− q20z2

+q40z4

+ · · · , and

s12 = O(z−1), s21 = O(z−1), s22 = eiν(t) +O(z−1),

Hence, we also obtain the asymptotic behavior (3.37) as z → ∞.

Moreover, on the one hand, substituting the WKB expansion µ±(λ;x, t) =n∑

i=−1

µ[i]± (x, t)zi + O

(zn+1

)( as z → 0)

into the time part of equivalent Lax pairs (3.16) and matching the O(λi)(i = −6,−5,−4,−3,−2,−1, 0) in order,combining the

O(1) :− 13

16iq100

[µ[2]± (x, t), σ3

]+

39

16q80σ3Q±

[µ[1]± (x, t), σ3

]+

39

16iq80[µ[0]± (x, t), σ3

]−

27

16q60σ3Q±

[µ[−1]± (x, t), σ3

]+ iσ3Q±µ

[−1]±,t (x, t) = iσ3Q±(T − T±)µ

[−1]± (x, t)

and(µ[−1]± (x, t)

)diag= 0, one can yield µ

[−1]±,t (x, t) 6= 0, ν±,t(x, t) 6= 0 and νt 6= 0 as x→ ±∞. That is, ν is dependent

on the variable t.

On the other hand, substituting the WKB expansion µ±(λ;x, t) =n∑i=0

µ[i]±

(x,t)

zi+O

(1

zn+1

)( as z → ∞) into the time

part of equivalent Lax pairs (3.16) and matching the O(λi)(i = 6, 5, 4, 3, 2, 1, 0) in order, combining the

O(1) :1

16i[µ[6]± (x, t), σ3

]+

1

16σ3Q±

[µ[5]± (x, t), σ3

]− 3

16iq20[µ[4]± (x, t), σ3

]−

13

16q20σ3Q±

[µ[3]± (x, t), σ3

]+

5

2iq40[µ[2]± (x, t), σ3

]+

5

2q40σ3Q±

[µ[1]± (x, t), σ3

]−

39

16iq60[µ[0]± (x, t), σ3

]+ µ

[0]±,t(x, t) = (T − T±)µ

[0]± (x, t)

and(µ[0]± (x, t)

)off= 0, one can yield µ

[0]±,t(x, t) 6= 0, ν±,t(x, t) 6= 0 and νt 6= 0 as x→ ±∞. That is, ν also is dependent

on the variable t. In summary, ν is a function about time t. Proof complete.�

3.1.5. Discrete Spectrum with Double Zeros.

In this section, we consider the case of s11(z) with double zeros and suppose that s11(z) has N1 double zeros inZ0 = {z ∈ C|Rez > 0, Imz > 0, |z| > q0} denoted by zn, and N2 double zeros on the circle W0 = {z ∈ C|z = q0e

iφ, 0 <φ < π

2 } denoted by ωm, that is, s11(z0) = s′11(z0) = 0 and s′′11(z0) 6= 0 if z0 is a double zero of s11(z). From the


symmetries of the scattering matrix presented in Proposition (16), the 4(2N1 + N2) discrete spectrum can be givenby the set

Z =

{± zn,±z∗n,±

q20zn,± q20

z∗n

}N1

n=1

⋃{± ωm,±ω∗

m

}N2

m=1, (3.39)

the distributions are shown in Figure 4. When a given z0 ∈ Z ∩D+, one can obtain that ψ+1(z0;x, t) and ψ−2(z0;x, t)are linearly dependent by combining Eq. (3.20) and s11(z0) = 0. Similarly, when a given z0 ∈ Z ∩D−, one can obtainthat ψ+2(z0;x, t) and ψ−1(z0;x, t) are linearly dependent by combining Eq. (3.20) and s22(z0) = 0. For convenience,we introduce the norming constant b[z0] such that

ψ+1(z0;x, t) = b[z0]ψ−2(z0;x, t), as z0 ∈ Z ∩D+,

ψ+2(z0;x, t) = b[z0]ψ−1(z0;x, t), as z0 ∈ Z ∩D−.(3.40)

For a given z0 ∈ Z ∩D+, according to s11(z0) =det(ψ+1(z;x,t),ψ−2(z;x,t)

)

ηin Eq. (3.20), and taking derivative respect

to z (here z = z0) on both sides of this equation, combining s11(z0) = 0, s′11(z0) = 0 and Eq. (3.40), we have

det(ψ′+1(z0;x, t), ψ−2(z0;x, t)) + det(ψ+1(z0;x, t), ψ

′−2(z0;x, t))

=det(ψ′+1(z0;x, t), ψ−2(z0;x, t)) + det(b[z0]ψ−2(z0;x, t), ψ

′−2(z0;x, t))

=det(ψ′+1(z0;x, t), ψ−2(z0;x, t))− det(b[z0]ψ

′−2(z0;x, t), ψ−2(z0;x, t))

=det(ψ′+1(z0;x, t)− b[z0]ψ

′−2(z0;x, t), ψ−2(z0;x, t)) = 0,

it is evidence that ψ′+1(z0;x, t) − b[z0]ψ

′−2(z0;x, t) and ψ−2(z0;x, t) are linearly dependent. Similarly, when a given

z0 ∈ Z∩D−, one can obtain that ψ′+2(z0;x, t)−b[z0]ψ′

−1(z0;x, t) and ψ−1(z0;x, t) are linearly dependent by combiningEq. (3.20) and s′22(z0) = 0. For convenience, we define another norming constant d[z0] such that

ψ′+1(z0;x, t)− b[z0]ψ

′−2(z0;x, t) = d[z0]ψ−2(z0;x, t), as z0 ∈ Z ∩D+,

ψ′+2(z0;x, t)− b[z0]ψ

′−1(z0;x, t) = d[z0]ψ−1(z0;x, t), as z0 ∈ Z ∩D−.

(3.41)

On the other hand, we notice that ψ+1(z;x, t) and s11(z) are analytic on D+. Suppose z0 is the double zeros ofs11(z), let ψ+1(z;x, t) and s11(z) carry out Taylor expansion at z = z0, we have

ψ+1(z;x, t)

s11(z)=

ψ+1(z0;x, t) + ψ′+1(z0;x, t)(z − z0) +

12!ψ

′′+1(z0;x, t)(z − z0)

2 + · · ·s11(z0) + s′11(z0)(z − z0) +

12!s

′′11(z0)(z − z0)2 +

13!s

′′′11(z0)(z − z0)3 + · · ·

=ψ+1(z0;x, t) + ψ′

+1(z0;x, t)(z − z0) +12!ψ

′′+1(z0;x, t)(z − z0)

2 + · · ·12!s

′′11(z0)(z − z0)2 +

13!s

′′′11(z0)(z − z0)3 + · · ·

=2ψ+1(z0;x, t)

s′′11(z0)(z − z0)

−2 +

(2ψ′

+1(z0;x, t)

s′′11(z0)− 2ψ+1(z0;x, t)s

′′′11(z0)

3s′′211(z0)

)(z − z0)

−1 + · · · ,

Then, one has the compact form

P−2z=z0

[ψ+1(z;x, t)

s11(z)

]=

2ψ+1(z0;x, t)

s′′11(z0)=

2b[z0]ψ−2(z0;x, t)

s′′11(z0), as z0 ∈ Z ∩D+,

where P−2z=z0

[f(z;x, t)] denotes the coefficient of O((z−z0)−2) term in the Laurent series expansion of f(z;x, t) at z = z0.

and

Resz=z0

[ψ+1(z;x, t)

s11(z)

]=

2ψ′+1(z0;x, t)

s′′11(z0)− 2ψ+1(z0;x, t)s

′′′11(z0)

3s′′211(z0)=

2(b[z0]ψ′−2(z0;x, t) + d[z0]ψ−2(z0;x, t))

s′′11(z0)−

2b[z0]ψ−2(z0;x, t)s′′′11(z0)

3s′′211(z0)=

2b[z0]ψ′−2(z0;x, t)

s′′11(z0)+

2d[z0]ψ−2(z0;x, t)

s′′11(z0)− 2b[z0]ψ−2(z0;x, t)s

′′′11(z0)

3s′′211(z0)

=2b[z0]ψ

′−2(z0;x, t)

s′′11(z0)+

2b[z0]

s′′11(z0)· d[z0]b[z0]

ψ−2(z0;x, t)−2b[z0]

s′′11(z0)· s

′′′11(z0)

3s′′11(z0)ψ−2(z0;x, t)

=2b[z0]ψ

′−2(z0;x, t)

s′′11(z0)+

[2b[z0]

s′′11(z0)

(d[z0]

b[z0]− s′′′11(z0)

3s′′11(z0)

)]ψ−2(z0;x, t), as z0 ∈ Z ∩D+,

where Resz=z0

[f(z;x, t)] denotes the coefficient of O((z−z0)−1) term in the Laurent series expansion of f(z;x, t) at z = z0.


Similarly, for the case of ψ+2(z;x, t) and s22(z) are analytic on D−, we repeat the above process and obtain

P−2z=z0

[ψ+2(z;x, t)

s22(z)

]=

2ψ+2(z0;x, t)

s′′22(z0)=

2b[z0]ψ−1(z0;x, t)

s′′22(z0), as z0 ∈ Z ∩D−,

Resz=z0

[ψ+2(z;x, t)

s22(z)

]=

2b[z0]ψ′−1(z0;x, t)

s′′22(z0)+

[2b[z0]

s′′22(z0)

(d[z0]

b[z0]− s′′′22(z0)

3s′′22(z0)

)]ψ−1(z0;x, t), as z0 ∈ Z ∩D−.

Moreover, let

A[z0] =

2b[z0]

s′′11(z0), as z0 ∈ Z ∩D+,

2b[z0]

s′′22(z0), as z0 ∈ Z ∩D−,

B[z0] =

d[z0]

b[z0]− s′′′11(z0)

3s′′11(z0), as z0 ∈ Z ∩D+,

d[z0]

b[z0]− s′′′22(z0)

3s′′22(z0), as z0 ∈ Z ∩D−.

(3.42)

Then, we have

P−2z=z0

[ψ+1(z;x, t)

s11(z)

]= A[z0]ψ−2(z0;x, t), as z0 ∈ Z ∩D+,

P−2z=z0

[ψ+2(z;x, t)

s22(z)

]= A[z0]ψ−1(z0;x, t), as z0 ∈ Z ∩D−,

Resz=z0

[ψ+1(z;x, t)

s11(z)

]= A[z0][ψ

′−2(z0;x, t) +B[z0]ψ−2(z0;x, t)], as z0 ∈ Z ∩D+,

Resz=z0

[ψ+2(z;x, t)

s22(z)

]= A[z0][ψ

′−1(z0;x, t) +B[z0]ψ−1(z0;x, t)], as z0 ∈ Z ∩D−.

(3.43)

Proposition 19. For a z0 ∈ Z, the three symmetry relations for A[z0] and B[z0] are deduced as follow• The first symmetry relation A[z0] = −A[z∗0 ]∗, B[z0] = B[z∗0 ]

∗.• The second symmetry relation A[z0] = A[−z∗0 ]∗, B[z0] = −B[−z∗0 ]∗.• The third symmetry relation A[z0] =

z40q∗−

q40q−A

[− q20

z0

], B[z0] =

q20z20B

[− q20

z0

]+ 2

z0.

Proof. For the first sysmetry relation, a given z0 ∈ Z, s11(z0) = s22(z0) = 0, hence substituting S(z0) =

(0 s12(z0)

s21(z0) 0

)

into ψ+(z0;x, t) = ψ−(z0;x, t)S(λ0), one can obtain

ψ+1(z0;x, t) = s21(z0)ψ−2(z0;x, t), as z0 ∈ Z ∩D+,

ψ+2(z0;x, t) = s12(z0)ψ−1(z0;x, t), as z0 ∈ Z ∩D−.

Therefore, when z0 ∈ Z ∩ D+ (z0 ∈ Z ∩ D−), one has obtain b[z0] = s21(z0) (b[z0] = s12(z0)). According toS(z0) = σ2S(z

∗0)

∗σ2 in Eq. (3.22), one can obtain

s11(z0) = s22(z∗0)

∗, s12(z0) = −s21(z∗0)∗, s21(z0) = −s12(z∗0)∗, s22(z0) = s11(z∗0)

∗, (3.44)

so we have b[z∗0 ]∗ = s12(z

∗0)

∗ = −s21(z0) = −b[z0] (where z∗0 ∈ Z ∩ D−), and A[z0] =2b[z0]s′′11(z0)

=−2b[z∗0 ]

∗

s′′22(z∗0 )

∗ = −A[z∗0 ]∗.In addition, for a given z0 ∈ Z ∩ D+ (z0 ∈ Z ∩ D−), by taking derivative respect to z (here z = z0) both sides ofψ+1(z0;x, t) = b[z0]ψ−2(z0;x, t) (ψ+2(z0;x, t) = b[z0]ψ−1(z0;x, t)), we have

ψ′+1(z0;x, t)− b[z0]ψ

′−2(z0;x, t) = b′[z0]ψ−2(z0;x, t), as z0 ∈ Z ∩D+,

ψ′+2(z0;x, t)− b[z0]ψ

′−1(z0;x, t) = b′[z0]ψ−1(z0;x, t), as z0 ∈ Z ∩D+.

Combining with Eq. (3.41), one can obtain

b[z0] = s21(z0) ⇒ d[z0] = b′[z0] = s′21(z0), as z0 ∈ Z ∩D+,

b[z0] = s12(z0) ⇒ d[z0] = b′[z0] = s′12(z0), as z0 ∈ Z ∩D−,

then lead to

B[z0] =d[z0]

b[z0]− s′′′11(z0)

3s′′11(z0)=s′21(z0)

s21(z0)− s′′′22(z

∗0)

∗

3s′′22(z∗0)

∗=

−s′12(z∗0)∗−s12(z∗0)∗

− s′′′22(z∗0)

∗

3s′′22(z∗0)

∗

=−d[z∗0 ]∗−b[z∗0 ]∗

− s′′′22(z∗0)

∗

3s′′22(z∗0)

∗=d[z∗0 ]

∗

b[z∗0 ]∗− s′′′22(z

∗0)

∗

3s′′22(z∗0)

∗= B[z∗0 ]

∗.

Similarly, by repeating the above process, we can demonstrate the second symmetry relation. Next, we will provethe third symmetry relation.


For a given z0 ∈ Z, s11(z0) = s22(z0) = 0, According to symmetry reduction S(z0) = (σ3Q−)−1S

(− q20z0

)σ3Q+ in

Eq. (3.24), one can obtain

(s11(z0) s12(z0)s21(z0) s22(z0)

)=

q∗+q∗−

s22

(− q20z0

)q+q∗−

s21

(− q20z0

)

q∗+q−s12

(− q20z0

)q+q−s11

(− q20z0

)

,

and

s11(z0) =q∗+q∗−s22

(− q20z0

)= 0, s12(z0) =

q+q∗−s21

(− q20z0

), s21(z0) =

q∗+q−s12

(− q20z0

), s22(z0) =

q+q−s11

(− q20z0

)= 0,

so we have b[− q20

z0

]= s12

(− q20

z0

)= q−

q∗+s21(z0) =

q−q∗+b[z0]

(where z0 ∈ Z ∩D+,− q20

z0∈ Z ∩D−

).

Due to the formula s11(z0) = s22

(− q20z0

)= 0, note that

s′11(z0) =q∗+q∗−s′22

(−q

20

z0

)· q

20

z20,

s′′11(z0) =q∗+q∗−s′′22

(−q

20

z0

)· q

40

z40+q∗+q∗−s′22

(−q

20

z0

)· (−2) · q

20

z30=q∗+q∗−s′′22

(−q

20

z0

)· q

40

z40, here s′22

(−q

20

z0

)= 0,

s′′′11(z0) =q∗+q∗−s′′′22

(−q

20

z0

)· q

60

z60+q∗+q∗−s′′22

(−q

20

z0

)· (−4) · q

40

z50+q∗+q∗−s′′22

(−q

20

z0

)· (−2) · q

40

z50+q∗+q∗−s′22

(−q

20

z0

)· 6 · q

20

z40

=q∗+q∗−s′′′22

(−q

20

z0

)· q

60

z60+q∗+q∗−s′′22

(−q

20

z0

)· (−4) · q

40

z50+q∗+q∗−s′′22

(−q

20

z0

)· (−2) · q

40

z50, here s′22

(−q

20

z0

)= 0,

we have

A[z0] =2b[z0]

s′′11(z0)=

2q∗+q−b[− q20z0

]

q∗+q∗−

q40z40s′′22

(− q20z0

) =z40q

∗−

q40q−

b[− q20z0

]

s′′22

(− q20z0

) =z40q

∗−

q40q−A

[−q

20

z0

].

Furthermore

d[z0] = b′[z0] = s

′

21(z0) =q∗+q−s′

12

(−q20z0

)·q20z20

=q∗+q−b′

[−q20z0

]·q20z20

=q∗+q−d

[−q20z0

]·q20z20, where z0 ∈ Z ∩D+

,−q20z0

∈ Z ∩D−,

so one can obtain

B[z0] =d[z0]

b[z0]− s′′′11(z0)

3s′′11(z0)=

q∗+q−d[− q20z0

]· q

20

z20

q∗+q−b[− q20z0

] −q∗+q∗−

s′′′22

(− q20z0

)q60z60

− 6q∗+q∗−

s′′22

(− q20z0

)q40z50

3q∗+q∗−

s′′22

(− q20z0

)q40z40

=q20z20

d[− q20z0

]

b[− q20z0

] −s′′′22

(− q20z0

)

3s′′22

(− q20z0

)

+

2

z0=q20z20B

[−q

20

z0

]+

2

z0.

This completes the proof. �

Proposition 20. For n = 1, 2, · · · , N1 and m = 1, 2, · · · , N2, we have

A[zn] = −A[z∗n]∗ = A[−z∗n]∗ = −A[−zn] =z4nq

∗−

q40q−A

[− q20zn

]=z4nq

∗−

q40q−A

[q20z∗n

]∗

= −z4nq

∗−

q40q−A

[− q20z∗n

]∗= −z

4nq

∗−

q40q−A

[q20zn

], zn ∈ Z0,

A[ωm] = −A[ω∗m]∗ = A[−ω∗

m]∗ = −A[−ωm] =

ω4mq

∗−

q40q−A[ωm]∗, ωm ∈W0,

B[zn] = B[z∗n]∗ = −B[−z∗n]∗ = −B[−zn] =

q20z2nB

[− q20zn

]+

2

zn=q20z2nB

[− q20z∗n

]∗+

2

zn

= − q20z2nB

[q20z∗n

]∗+

2

zn= − q20

z2nB

[q20zn

]+

2

zn, zn ∈ Z0,

B[ωm] = B[ω∗m]∗ = −B[−ω∗

m]∗ = −B[−ωm] = − q20

ω2m

B[ωm]∗ +2

ωm, ωm ∈W0.

(3.45)


Proof. From the proposition (19), the symmetry relation A[zn] = −A[−zn] can be derived by combining A[z0] =

−A[z∗0 ]∗ and A[z0] = A[−z∗0 ]∗, and we directly obtain A[zn] = −A[z∗n]∗ = A[−z∗n]∗ =z40q

∗−

q40q−A[− q20zn

]and

z40q∗−

q40q−A[− q20zn

]=

z40q∗−

q40q−A[q20z∗n

]∗= − z40q

∗−

q40q−A[− q20z∗n

]∗= − z40q

∗−

q40q−A[q20zn

]. In conclusion, we obtain the following identity

A[zn] = −A[z∗n]∗ = A[−z∗n]∗ = −A[−zn] =z4nq

∗−

q40q−A

[− q20zn

]=z4nq

∗−

q40q−A

[q20z∗n

]∗

= −z4nq

∗−

q40q−A

[− q20z∗n

]∗= −z

4nq

∗−

q40q−A

[q20zn

], zn ∈ Z0.

Furthermore, here let zn = ωm = q0eiφ, substituting it into A[zn] =

z4nq∗−

q40q−A

[− q20

zn

], one can obtain

A[ωm] =ω4mq

∗−

q40q−A

[− q0e

−iφ

]=ω4mq

∗−

q40q−A[−ω∗

m] =ω4mq

∗−

q40q−A[ωm]∗, ωm ∈ W0,

and

A[ωm] = −A[ω∗m]∗ = A[−ω∗

m]∗ = −A[−ωm] =

ω4mq

∗−

q40q−A[ωm]∗, ωm ∈ W0.

Similarly, the identities about B[zn] and B[ωm] can be derived by repeating above process. End of proof.�

3.2. Inverse Scattering Problem with NZBCs and Double Poles.

In the following subsections, we will propose an inverse problem with NZBCs and solve it to obtain explicit doublepoles solutions for the TOFKN equation (1.1).

3.2.1. The Matrix Riemann-Hilbert Problem with NZBCs and Double Poles.

Similar to the case of ZBCs in Section 2, the matrix RH problem for the NZBCs can also be established. In order

to pose and solve the RH problem conveniently, we define ζn = − q20ζn

with

ζn =

zn, n = 1, 2, · · · , N1,

− zn−N , n = N1 + 1, N1 + 2, · · · , 2N1,

q20z∗n−2N1

, n = 2N1 + 1, 2N1 + 2, · · · , 3N1,

− q20z∗n−3N1

, n = 3N1 + 1, 3N1 + 2, · · · , 4N1,

ωn−4N1 , n = 4N1 + 1, 4N1 + 2, · · · , 4N1 +N2,

− ωn−4N1−N2 , n = 4N1 +N2 + 1, 4N1 +N2 + 2, · · · , 4N1 + 2N2.

(3.46)

Then, we can proposed a matrix RH problem as follows.

Proposition 21. Define the sectionally meromorphic matrices

M(z;x, t) =

M+(z;x, t) =

(µ+1(z;x, t)

s11(z), µ−2(z;x, t)

), as z ∈ D+,

M−(z;x, t) =

(µ−1(z;x, t),

µ+2(z;x, t)

s22(z)

), as z ∈ D−,

(3.47)

where limz′→z

z′∈D±

M(z′;x, t) =M±(z;x, t). Then, the multiplicative matrix Riemann-Hilbert problem is given below:

• Analyticity: M(z;x, t) is analytic in D+∪D−\Z and has the double poles in Z, whose principal parts of the Laurent

series at each double pole ζn or ζn, are determined as

Resz=ζn

M(z;x, t) =(A[ζn]e


),

P−2z=ζn

M(z;x, t) =(A[ζn]e

−2iθ(ζn;x,t)µ−2(ζn;x, t), 0),

Resz=ζn

M(z;x, t) =(0, A[ζn]e

2iθ(ζn;x,t){µ′−1(ζn;x, t) + [B[ζn] + 2iθ′(ζn;x, t)]µ−1(ζn;x, t)}

),

P−2

z=ζn

M(z;x, t) =(0, A[ζn]e

2iθ(ζn;x,t)µ−1(ζn;x, t)).

(3.48)


• Jump condition:

M−(z;x, t) =M+(z;x, t)[I − J(z;x, t)], as z ∈ Σ, (3.49)

where

J(z;x, t) = eiθ(z;x,t)σ3

(0 −ρ(z)ρ(z) ρ(z)ρ(z)

). (3.50)

• Asymptotic behavior:

M(z;x, t) =

i

zeiν−(x,t)σ3σ3Q− +O(1), as z → 0,

eiν−(x,t)σ3 +

(1

z

), as z → ∞.

(3.51)

Proof. For the analyticity of M(z;x, t), It follows from Eqs. (3.13) and (3.43) that for each double poles ζn ∈ D+ or

ζn ∈ D−. Now, we consider ζn ∈ D+, and obtain

Resz=ζn

[ψ+1(ξn;x, t)

s11(ζn)

]= Res

z=ζn

[µ+1(ζn;x, t)


]= A[ζn][ψ

′−2(ζn;x, t) +B[ζn]ψ−2(ζn;x, t)]

= A[ζn]{[µ−2(ζn;x, t)e−iθ(ζn;x,t)]′ +B[ζn]µ−2(ζn;x, t)e

−iθ(ζn;x,t)},⇒

Resz=ζn

[µ+1(ζn;x, t)

s11(ζn)

]= A[ζn]e

−2iθ(ζn;x,t){µ′−2(ζn;x, t) + [B[ζn]− 2iθ′(ζn;x, t)]µ−2(ζn;x, t)},⇒

Resz=ζn

M(z;x, t) =(A[ζn]e


),

where the “′” denotes the partial derivative with respect to z (here z = ζn), and

P−2z=ζn

[ψ+1(ζn;x, t)

s11(ζn)

]= P−2z=ζn

[µ+1(ζn;x, t)


]= A[ζn]µ−2(ζn;x, t)e

−iθ(ζn;x,t),⇒

P−2z=ζn

[µ+1(ζn;x, t)

s11(ζn)

]= A[ζn]e

−2iθ(ζn;x,t)µ−2(ζn;x, t),⇒

P−2λ=ζn


−2iθ(ζn;x,t)µ−2(ζn;x, t), 0).

Similarly, we also can obtain the analyticity for ζn ∈ D−. It follows from Eqs. (3.13) and (3.19) that

{µ+1(z;x, t) = µ−1(z;x, t)s11(z) + µ−2(z;x, t)e

−2iθ(z;x,t)s21(z),

µ+2(z;x, t) = µ−1(z;x, t)e2iθ(z;x,t)s12(z) + µ−2(z;x, t)s22(z),

by combining formula (3.21), one can obtain

M+(z;x, t) =

(µ+1(z;x, t)

s11(z), µ−2(z;x, t)

)=(µ−1(z;x, t) + µ−2(z;x, t)e

−2iθ(z;x,t)ρ(z), µ−2(z;x, t)),

M−(z;x, t) =

(µ−1(z;x, t),

µ+2(z;x, t)

s22(z)

)=(µ−1(z;x, t), µ−1(z;x, t)e

2iθ(z;x,t)ρ(z) + µ−2(z;x, t)),

and

(µ−1(z;x, t),

µ+2(z;x, t)

s22(z)

)=

(µ+1(z;x, t)

s11(z), µ−2(z;x, t)

)(1 e2iθ(z;x,t)ρ(z)

−e−2iθ(z;x,t)ρ(z) 1− ρ(z)ρ(z)

),

M−(z;x, t) =M+(z;x, t)(I − J(z;x, t)),

where J(z;x, t) is given by Eq. (3.50). The asymptotic behaviors of the modified Jost solutions µ±(z;x, t) andscattering matrix S(z) given in Propositions (17) and (18) can easily lead to the asymptotic behavior of M(z;x, t).


Specifically, when z → 0, we have

M+(z;x, t) =

(µ+1(z;x, t)

s11(z), µ−2(z;x, t)

)=

0 izq−e

iν−(x,t)

iq∗+e−iν+(x,t)

zq−q+

eiν(t)0

+O(1)

=i

z

(0 q−e

iν−(x,t)

q∗−e−iν−(x,t) 0

)+O(1) =

i

zeiν−(x,t)σ3σ3Q− +O(1),

M−(z;x, t) =

(µ−1(z;x, t),

µ+2(z;x, t)

s22(z)

)=

0

izq+eiν+(x,t)

q+q−

e−iν(t)

izq∗−e

−iν−(x,t) 0

+O(1)

=i

z

(0 q−e

iν−(x,t)

q∗−e−iν−(x,t) 0

)+O(1) =

i

zeiν−(x,t)σ3σ3Q− +O(1),

it can obtain that M(z;x, t) = izeiν−(x,t)σ3σ3Q− +O(1) as z → 0. Moreover, when z → ∞, we have

M+(z;x, t) =

(µ+1(z;x, t)

s11(z), µ−2(z;x, t)

)=

(eiν+(x,t)

e−iν(t) 0

0 e−iν−(x,t)

)+O

(1

z

)

=

(eiν−(x,t) 0

0 e−iν−(x,t)

)+O

(1

z


(1

z

),

M−(z;x, t) =

(µ−1(z;x, t),

µ+2(z;x, t)

s22(z)

)=

(eiν−(x,t) 0

0 e−iν+(x,t)

eiν(t)

)+O

(1

z

)

=

(eiν−(x,t) 0

0 e−iν−(x,t)

)+O

(1

z


(1

z

),

it can obtain that M(z;x, t) = eiν−(x,t)σ3 + O(1z

). Therefore, we obtain the asymptotic behavior (3.51) of M(z;x, t)

when z → 0 and z → ∞. This completes the proof.�

Proposition 22. The solution of the matrix Riemann-Hilbert problem with double poles can be expressed as

M(z;x, t) =eiν−(x,t)σ3

(I +

i

zσ3Q−

)+

1

2πi

∫

Σ

M+(ξ;x, t)J(ξ;x, t)

ξ − zdξ+

4N1+2N2∑

n=1

(Cn(z)

[µ′−2(ζn;x, t) +

(Dn +

1

z − ζn

)µ−2(ζn;x, t)

],

Cn(z)

[µ′−1(ζn;x, t) +

(Dn +

1

z − ζn

)µ−1(ζn;x, t)

]),

(3.52)

where∫Σstands for an integral along the oriented contour displayed in Fig. 4,

Cn(z) =A[ζn]

z − ζne−2iθ(ζn;x,t), Cn(z) =

A[ζn]

z − ζne2iθ(ζn;x,t),

Dn = B[ζn]− 2iθ′(ζn;x, t), Dn = B[ζn] + 2iθ′(ζn;x, t),

(3.53)

µ−2(ζn;x, t) and µ′−2(ζn;x, t) are determined via µ−1(ζn;x, t) and µ

′−1(ζn;x, t) as

µ−2(ζn;x, t) =iq−ζn

µ−1(ζn;x, t), µ′−2(ζn;x, t) = − iq−

ζ2nµ−1(ζn;x, t) +

iq−q20

ζ3nµ′−1(ζn;x, t), (3.54)


and µ−1(ζn;x, t) and µ′−1(ζn;x, t) satisfy the linear system of 8N1 + 4N2 as below:

4N1+2N2∑

n=1

{Cn(ζs)µ

′−1(ζn;x, t) +

[Cn(ζs)

(Dn +

1

ζs − ζn

)− iq−

ζsδs,n

]µ−1(ζn;x, t)

}=

− eiν−(x,t)σ3

(iq−ζs

1

)− 1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))2ξ − ζs

dξ,

4N1+2N2∑

n=1

{(Cn(ζs)

ζs − ζn+iq−q

20

ζ3sδs,n

)µ′−1(ζn;x, t) +

[Cn(ζs)

ζs − ζn

(Dn +

2

ζs − ζn

)− iq−

ζ2sδs,n

]µ−1(ζn;x, t)

}=

− eiν−(x,t)σ3

(iq−ζ2s

0

)+

1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))2(ξ − ζs)2

dξ,

(3.55)

where s = 1, 2, · · · , 4N1 + 2N2 and δs,n are the Kronecker δ-symbol.

Proof. Similar to the proof of Proposition (9) for the case of ZBCs. In order to regularize the RH problem establishedin Proposition (21) for the case of NZBCs, one has to subtract out the asymptotic values as z → 0 and z → ∞ givenby Eq. (3.51) and the singular contributions. Then, the jump condition (3.49) becomes

M−(z;x, t)− i

zeiν−(x,t)σ3σ3Q− − eiν−(x,t)σ3−

4N1+2N2∑

n=1

P−2z=ζn

M(z;x, t)

(z − ζn)2+

Resz=ζn

M(z;x, t)

z − ζn+

P−2

z=ζn

M(z;x, t)

(z − ζn)2+

Resz=ζn

M(z;x, t)

z − ζn

=M+(z;x, t)− i

zeiν−(x,t)σ3σ3Q−−

eiν−(x,t)σ3 −4N1+2N2∑

n=1

P−2z=ζn

M(z;x, t)

(z − ζn)2+

Resz=ζn

M(z;x, t)

z − ζn+

P−2

z=ζn

M(z;x, t)

(z − ζn)2+

Resz=ζn

M(z;x, t)

z − ζn

−M+(z;x, t)J(z;x, t),

(3.56)

where P−2z=ζn

M(z;x, t), Resz=ζn

M(z;x, t), P−2

z=ζn

M(z;x, t), Resz=ζn

M(z;x, t) have given in Eq. (3.48). By using Plemelj’s for-

mula, one can obtain the solution (3.52) with formula (3.53) of the matrix RH problem. According to the symmetry

reduction µ±(z;x, t) =izµ±

(− q20

z;x, t

)σ3Q±, we can obtain

µ−2(z;x, t) =i

zq−µ−1

(− q20

z;x, t

),

and take derivative respect to z on both sides of above equation at the same time. Then, let z = ζn, one have followingformula

µ′−2(ζn;x, t) = − iq−

ζ2nµ−1(ζn;x, t) +

iq−q20

ζ3nµ′−1(ζn;x, t).

From Eq. (3.47), when we take z = ζs, µ−2(ζs;x, t) is the second column element of matrix M(ζs;x, t), that is

µ−2(ζs;x, t) =

(iζseiν−(x,t)q−e−iν−(x,t)

)+

1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))2ξ − ζs

dξ+

4N1+2N2∑

n=1

{Cn(ζs)

[µ′−1(ζn;x, t) +

(Dn +

1

ζs − ζn

)µ−1(ζn;x, t)

]},

(3.57)

where s = 1, 2, · · · , 4N1 + 2N2 and ζs is equivalent to ζn. Then, taking the derivative respect to z (here z = ζs) onboth sides of Eq. (3.57) at the same time, we have

µ′−2(ζs;x, t) =

(− iζ2sq−e

iν−(x,t)

0

)+

1

2πi

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))2(ξ − ζs)2

dξ−

4N1+2N2∑

n=1

Cn(ζs)

{1

ζs − ζn

[µ′−1(ζn;x, t) +

(Dn +

2

ζs − ζn

)µ−1(ζn;x, t)

]}.

(3.58)

After that, substituting Eqs. (3.57)-(3.58) into Eq. (3.54) to eliminate µ−2(ζs;x, t) and µ′−2(ζs;x, t), by merging

and simplifying, one can derived Eq. (3.55). Completing the proof.�


3.2.2. Reconstruction Formula of the Potential.

From the solution (3.52) of the matrix RH problem, we have

M(z;x, t) = eiν−(x,t)σ3 +M [1](x, t)

z+O

(1

z2

), as z → ∞, (3.59)

where

M[1](x, t) =ieiν−(x,t)σ3σ3Q− −

1

2πi

∫

Σ

M+(ξ;x, t)J(ξ;x, t)dξ +

4N1+2N2∑

n=1

{A[ζn]e

−2iθ(ζn ;x,t)[µ′

−2(ζn;x, t) +Dnµ−2(ζn;x, t)],

A[ζn]e2iθ(ζn ;x,t)

[µ′

−1(ζn; x, t) + Dnµ−1(ζn;x, t)]}.

(3.60)

Substituting Eq. (3.59) into Eq. (3.15) and matching O(z) term, we have

O(z) :i

2

[M [1](x, t), σ3

]+

1

2

[σ3Q±e

iν−(x,t)σ3 , σ3]= (Q−Q±)e

iν−(x,t)σ3 ,

then, by expanding the above equation, one can find the reconstruction formula of the double poles solution (potential)for the TOFKN equation with NZBCs as follows

q(x, t) = −ieiν−(x,t)M[1]12 (x, t), (3.61)

where M[1]12 (x, t) represents the first row and second column element of the matrix M [1](x, t), and

M[1]12 (x, t) = ieiν−(x,t)

q− −1

2πi

∫

Σ

[M

+(ξ;x, t)J(ξ;x, t)]12dξ +

4N1+2N2∑

n=1

{A[ζn]e

2iθ(ζn ;x,t)[µ′

−11(ζn;x, t) + Dnµ−11(ζn;x, t)]},

(3.62)

when taking row vector α =(α(1), α(2)

)and column vector γ = (γ(1), γ(2))T, where

α(1) =

(A[ζn]e

2iθ(ζn;x,t)

)

1×(4N1+2N2)

, α(2) =

(A[ζn]e

2iθ(ζn;x,t)Dn

)

1×(4N1+2N2)

,

γ(1) =

(µ′−11(ζn;x, t)

)

1×(4N1+2N2)

, γ(2) =

(µ−11(ζn;x, t)

)

1×(4N1+2N2)

,

(3.63)

we can obtain a more concise reconstruction formulation of the double-pole solution (potential) for the TOFKNequation with NZBCs as follows

q(x, t) = eiν−(x,t)

[q−e

iν−(x,t) − iαγ +1

2π

∫

Σ

(M+(ξ;x, t)J(ξ;x, t))12dξ

]. (3.64)

3.2.3. Trace Formulae and Theta Condition.

The so-called trace formulae are that the scattering coefficients s11(z) and s22(z) are formulated in terms of thediscrete spectrum Z and reflection coefficients ρ(z) and ρ(z). We know that s11(z), s22(z) are analytic on D+, D−,

respectively. The discrete spectral points ζn’s are the double zeros of s11(z), while ζn’s are the double zeros of s22(z).Define the functions β±(z) as follows:

β+(z) = s11(z)

4N1+2N2∏

n=1

(z − ζnz − ζn

)2

eiν ,

β−(z) = s22(z)

4N1+2N2∏

n=1

(z − ζn

z − ζn

)2

e−iν .

(3.65)

Then, β+(z) and β−(z) are analytic and have no zero in D+ and D−, respectively. Furthermore, we have therelation β+(z)β−(z) = s11(z)s22(z) and the asymptotic behaviors β±(z) → 1, as z → ∞.

According to detS(z) = s11s22 − s21s12 = 1, we can derive

1

s11(z)s22(z)= 1− s21(z)s12(z)

s11(z)s22(z)= 1− ρ(z)ρ(z),

by taking logarithm on both sides of the above equation at the same time, we have

−log(s11(z)s22(z)) = log[1− ρ(z)ρ(z)] ⇒ log[β+(z)β−(z)] = −log[1− ρ(z)ρ(z)],

then employing the Plemelj’ formula such that one has

logβ±(z) = ∓ 1

2πi

∫

Σ

log[1− ρ(z)ρ(z)]

ξ − zdξ, z ∈ D±. (3.66)


Substituting Eq. (3.66) into Eq. (3.65), we can obtain the trace formulae

s11(z) = exp

(− 1

2πi

∫

Σ


ξ − zdξ

)4N1+2N2∏

n=1

(z − ζn

z − ζn

)2

e−iν ,

s22(z) = exp

(1

2πi

∫

Σ


ξ − zdξ

)4N1+2N2∏

n=1

(z − ζnz − ζn

)2

eiν .

(3.67)

As z → 0, we consider formulas (3.67) and the asymptotic behavior of the scattering matrix (3.37), then theso-called theta condition is obtained. That is to say, there exists i ∈ Z such that

arg

(q−q+

)+ 2ν = 16

N1∑

n=1

arg(zn) + 8

N2∑

n=1

arg(ωm) + 2πi+1

2π

∫

Σ

log[1− ρ(ξ)ρ(ξ)]

ξdξ. (3.68)

3.2.4. Reflectionless Potential: Double-Pole Soliton Solutions.

For the case of the reflectionless potential: ρ(z) = ρ(z) = 0, the part jump matrix J(z;x, t) in Eq. (3.50) canbe simplified as J(z;x, t) = 02×2. From the Volterra integral equation (72), one can derive ψ±(q0;x, t) = Y±q0.Combining with the definition of scattering matrix, one has S(q0) = I and q+ = q−. From the theta condition, thereexists i ∈ Z lead to

ν = 8

N1∑

n=1

arg(zn) + 4

N2∑

n=1

arg(ωm) + πi. (3.69)

Then Eqs. (3.55) and (3.64) with J(z;x, t) = 02×2 become

4N1+2N2∑

n=1

{Cn(ζs)µ

′−1(ζn;x, t) +

[Cn(ζs)

(Dn +

1

ζs − ζn

)− iq−

ζsδs,n

]µ−1(ζn;x, t)

}=

− eiν−(x,t)σ3

(iq−ζs

1

),

4N1+2N2∑

n=1

{(Cn(ζs)

ζs − ζn+iq−q

20

ζ3sδs,n

)µ′−1(ζn;x, t) +

[Cn(ζs)

ζs − ζn

(Dn +

2

ζs − ζn

)− iq−

ζ2sδs,n

]µ−1(ζn;x, t)

}=

− eiν−(x,t)σ3

(iq−ζ2s

0

),

(3.70)

and

q(x, t) = eiν−(x,t)(q−e

iν−(x,t) − iαγ). (3.71)

Theorem 23. The explicit expression for the double-pole solution of the TOFKN equation (1.1) with NZBCs is givenby determinant formula

q(x, t) = q−

(det(R)

det(G)

)2det(G)

det(R)

(1 +

det(G)

det(R)

), (3.72)

where

R =

(0 ατ G

), R =

(0 α

τ G

), τ =

(τ (1)

τ (2)

),

τ (1) =

(1

ζs

)

(4N1+2N2)×1

, τ (2) =

(1

ζ2s

)

(4N1+2N2)×1

,

(3.73)

and the (8N1+4N2)×(8N1+4N2) partitioned matrix G =

(G(1,1) G(1,2)

G(2,1) G(2,2)

)with G(i,j) =

(g(i,j)s,n

)

(4N1+2N2)×(4N1+2N2)

(i, j = 1, 2) given by

g(1,1)s,n = Cn(ζs), g(1,2)s,n = Cn(ζs)

(Dn +

1

ζs − ζn

)− iq−

ζsδs,n,

g(2,1)s,n =Cn(ζs)

ζs − ζn+iq−q

20

ζ3sδs,n, g(2,2)s,n =

Cn(ζs)

ζs − ζn

(Dn +

2

ζs − ζn

)− iq−

ζ2sδs,n,


and the (8N1+4N2)×(8N1+4N2) partitioned matrix G =

(G(1,1) G(1,2)

G(2,1) G(2,2)

)with G(i,j) =

(g(i,j)s,n

)

(4N1+2N2)×(4N1+2N2)

(i, j = 1, 2) given by

g(1,1)s,n =λ(ζs)

λ(ζn)Cn(ζs), g(1,2)s,n =

λ(ζs)

λ(ζn)Cn(ζs)

(Dn +

1

ζs − ζn− λ′(ζn)

λ(ζn)

)− iq−

ζsδs,n,

g(2,1)s,n =Cn(ζs)

λ(ζn)

(λ(ζs)

ζs − ζn− λ′(ζs)

)+iq−q

20

ζ3sδs,n,

g(2,2)s,n =

[λ(ζs)

ζs − ζn

(Dn +

2


λ(ζn)

)− λ′(ζs)

(Dn +

1


λ(ζn)

)]− iq−

ζ2sδs,n.

Proof. From the Eqs. (3.63), (3.70) and (3.71), the reflectionless potential is deduced by determinants:

q(x, t) = q−e2iν−(x,t)

(1 +

det(R)

det(G)

). (3.74)

However, this formula (3.74) is implicit since ν−(x, t) is included. One needs to derive an explicit form for thereflectionless potential. From the trace formulae (3.67) and Volterra integral equation (3.18) as x→ −∞, one derivesthat

M(z;x, t) =Y−(z) + λ(z)

4N1+2N2∑

n=1

P−2z=ζn

(M(λ(z);x, t)/λ(z)

)

(z − ζn)2+

Resz=ζn

(M(λ(z);x, t)/λ(z)

)

z − ζn+

P−2

z=ζn

(M(λ(z);x, t)/λ(z)

)

(z − ζn)2+

Resz=ζn

(M(λ(z);x, t)/λ(z)

)

z − ζn

,

(3.75)

which can yield the γ given by Eq. (3.63) exactly. Then, substituting γ into the formula of the potential, one yields

q(x, t) = q−eiν−(x,t)

(eiν−(x,t) +

det(R)

det(G)

), (3.76)

then, by combining Eq. (3.74) with Eq. (3.76), we can obtain Eq. (3.72), and complete the proof.�

For example, we exhibit the double-pole solutions which contain three kind of types for the TOFKN equation withNZBCs:

• When taking parameters N1 = 0, N2 = 1, q± = 1, ω1 = eπ4 i, A[ω1] = i, B[ω1] = 1 + (1 −

√2)i, we can obtain the

explicit double-pole dark-bright soliton solution q(x, t) = E1

E2with

E1 =− i{(3√2it+ 3

√2t+

√2− 1− i)2

(e2x+7t

)2+[2(12

√2it − 36it2 + 2

√2i− 12it− 4i)(3

√2it+ 3

√2t+

√2− 1− i)

ex+72t + 2(−3

√2it −

√2i+ 3

√2t− 1 + i)(3

√2it+ 3

√2t+

√2− 1− i)

]e2x+7t + (12

√2it − 36it2 + 2

√2i− 12it− 4i)2

(ex+

72t)2

+ 2(−3√2it−

√2i+ 3

√2t− 1 + i)(12

√2it− 36it2 + 2

√2i− 12it − 4i)ex+

72t + (−3

√2it−

√2i+ 3

√2t− 1

+i)2} [

(12√2it− 36it2 + 2

√2i− 12it + 4

√2− 24t − 4)e2x+7t + (2

√2i− 6

√2t− 2

√2 + 2)ex+

72t + (2

√2i− 6

√2t+ 2)

e3x+212

t + ie4x+14t + i],

E2 =[(12

√2t− 36t2 + 2

√2− 12t− 4)e2x+7t + (−6

√2t− 2

√2 + 2)ex+

72t + (6

√2t− 2)e3x+

212

t − e4x+14t − 1]

[(3√2it+

√2i− 3

√2t+ 1− i)e2x+7t + (12

√2it− 36it2 + 2

√2i− 12it + 2

√2− 2− 2i− 12t)ex+

72t−

3√2it− 3

√2t−

√2 + 1 + i

]2,

and give out relevant plots in Fig. 5. Fig. 5 (a) and (b) exhibit the three-dimensional and density diagrams for theexact double-pole dark-bright soliton solution of the TOFKN equation with NZBCs. Fig. 5 (c) displays the distinctprofiles of the exact double-pole soliton for t = ±3, 0. It is a semi rational soliton, which is different from the simplepole solution usually expressed by exponential function, even if the double-pole dark-bright soliton solution shows theinteraction of dark soliton and bright soliton.

Compared with the classical second-order flow Kaup-Newell system which also be called the DNLS equation inRef. [44], according to the density diagrams of the double-pole dark-bright soliton solutions for the TOFKN equationand the DNLS equation in Ref. [44] shows that the trajectories of solutions are different obviously. Moreover, by


(a)-15 -10 -5 0 5 10 15

x

-10

-8

-6

-4

-2

0

2

4

6

8

10

t

(b) (c)

Figure 5. (Color online) The double-pole dark-bright soliton solution of TOFKN equation (1.1)

with NZBCs and N1 = 0, N2 = 1, q± = 1, ω1 = eπ4 i, A[ω1] = i, B[ω1] = 1 + (1 −

√2)i. (a) The

three-dimensional plot; (b) The density plot; (c) The sectional drawings at t = −3 (dashed line),t = 0 (solid line), and t = 3 (dash-dot line).

(a)-4 -3 -2 -1 0 1 2 3 4 5

x

-3

-2

-1

0

1

2

3

t

(b) (c)

Figure 6. (Color online) The double-pole breather-breather solution of TOFKN equation (1.1) withNZBCs and N1 = 1, N2 = 0, q± = 1, ζ1 = 2e

π4 i, A[ζ1] = i, B[ζ1] = i. (a) The three-dimensional plot;

(b) The density plot; (c) The sectional drawings at t = −5 (dashed line), t = 0 (solid line), and t = 5(dash-dot line).

observing the form of exact double-pole dark-bright soliton solution, the average wave velocity of the double-poledark-bright soliton solution of the DNLS equation from Ref. [44] is about −2, while the average wave velocity of thedouble-pole dark-bright soliton solution in this paper is about −7/2. It further validates the introduction of third-orderdispersion and quintic nonlinear term of Kaup-Newell systems can affect the trajectories and the speed of solutions.

Remark 24. The parameter selection of the double-pole dark-bright soliton solution shown in Fig. 5. is consistentwith that in reference [44].

• When taking parameters N1 = 1, N2 = 0, q± = 1, ζ1 = 2eπ4 i, A[ζ1] = i, B[ζ1] = i, we can obtain the explicit

double-pole breather-breather solution and give out relevant plots in Fig. 6. Figs. 6 (a) and (b) exhibit the three-dimensional and density diagrams for the exact double-pole breather-breather solution of the TOFKN equation withNZBCs. Fig. 6 (c) displays the distinct profiles of the exact double-pole breather-breather solution for t = ±5, 0.Moreover, from the density plot Fig. 6 (b), we can find that the propagation of the double-pole breather-breathersolution is localized in space x and periodic in time t. Obviously, from the Fig. 6 (a) and (b), we can find that thedouble-pole breather-breather solution breathes 10 times at t = [−3, 3] and collides once at near t = 0.

• When taking parameters N1 = 1, N2 = 1, q± = 1, ζ1 = 2eπ4 i, ω1 = e

π4 i, A[ζ1] = B[ζ1] = i, A[ω1] = B[ω1] = i,

we can obtain the explicit double-pole breather-breather-dark-bright solution and give out relevant plots in Fig. 7.Figs. 7 (a) and (b) exhibit the three-dimensional and density diagrams for the exact double-pole breather-breather-dark-bright solution of the TOFKN equation with NZBCs, which is equivalent to a combination of breather-breathersolution and dark-bright soliton solutions. Fig. 7 (c) displays the distinct profiles of the exact double-pole breather-breather-dark-bright solution for t = ±5, 0. It is obvious that the double-pole breather-breather-dark-bright solutionis an interaction between the double-pole breather-breather solution and the double-pole dark-bright soliton solution.Moreover, We can find that the amplitude at t = 0 in Fig. 7 (c) is between that the amplitude at t = 0 in Fig. 5 (c)and the amplitude at t = 0 in Fig. 6 (c), which also in accordance with the energy conservation law of the interactionbetween two waves.


(a)-15 -10 -5 0 5 10 15

x

-3

-2

-1

0

1

2

3

t

(b) (c)

Figure 7. (Color online) The double-pole breather-breather-dark-bright solution of TOFKN equa-tion (1.1) with NZBCs and N1 = 1, N2 = 1, q± = 1, ζ1 = 2e

π4 i, ω1 = e

π4 i, A[ζ1] = B[ζ1] = i, A[ω1] =

B[ω1] = i. (a) The three-dimensional plot; (b) The density plot; (c) The sectional drawings at t = −5(dashed line), t = 0 (solid line), and t = 5 (dash-dot line).

4. Conclusions and Discussions

In this paper, we reveal the corresponding parameter reduction from the coupled TOFKN equation to the generalform of the TOFKN equation (1.1) specifically. the inverse scattering transforms for the TOFKN equation underZBCs and NZBCs have been built by applying RH method. We not only construct the direct scattering problemwhich illustrates the analyticity, symmetries, asymptotic behaviors and discrete spectrum, but also establish and solvethe inverse problem which can derive the trace formula and reflectionless potential with the aid of a matrix RH problem.For the case of ZBCs, from the asymptotic behavior for the scattering matrix in the direct scattering problem, weconnect the constant ν with the conservation of mass (also called power) I1 of the modified Zakharov-Shabat spectralproblem and exhibit essential proof in details. Due to the multi-valued case of eigenvalue k(λ) in the case of NZBCs,we illustrate the scattering problem on a standard z-plane by utilizing two single-valued inverse mappings. Since theintroduction of z-plane, the inverse scattering transform with NZBCs is more complex than the case of ZBCs. Inparticular, the reflectionless potentials with double poles which also be called the double-pole solutions of the TOFKNequation (1.1) are deduced explicitly via the determinants. Based on the RH method, we further improve the rigoroustheory of inverse scattering transforms of higher-order flow KN system, and also provides a valuable reference for theinverse scattering transforms of the higher-order flow equations of KN systems or other nonlinear integrable dynamicalsystems.

Vivid illustration and dynamic behavior for some representative double-pole solutions have been given out indetails. In the case of ZBCs, we exhibit the dynamic behaviors of the N -double-pole solution when taking N = 1and N = 2 respectively, and discover the exact 1-double-pole soliton solution is equivalent to the elastic collisions oftwo bright-bright solitons. Furthermore, the exact 2-double-pole soliton solution is equivalent to the interaction oftwo 1-double-pole soliton solutions. In the case of NZBCs, when N1 and N2 take different values, we give out thedynamic behaviors of abundant double-pole solutions which contain the double-pole dark-bright soliton solution asN1 = 0, N2 = 1, the double-pole breather-breather solution as N1 = 1, N2 = 0 and the double-pole breather-breather-dark-bright solution as N1 = 1, N2 = 1.

In addition, by comparing the solutions of the TOFKN equation and the DNLS equation, it is found that thethird-order dispersion and quintic nonlinear term of the KN system can affect both the trajectory and the speed ofthe solutions. However, the third-order dispersion and quintic nonlinear term of KN system have little influence onthe maximum amplitude of the solution at a certain moment. These conclusions are consistent with those in Ref. [46].These analytical results obtained in this work might have vital reference value for the study of the higher-order flowequations of KN systems or other nonlinear integrable dynamical systems, and provide a theoretical basis for possibleexperimental research and applications. At present, there are more and more researches on the triple poles solutionsand even the N-pole solutions, and the long-time asymptotic behaviors have attracted more and more attention. In thefuture, we will study the multipole solutions and long-time asymptotic behaviors for the higher-order flow equationsof KN systems or other nonlinear integrable dynamical systems.

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(JP) School of Mathematical Sciences, Shanghai Key Laboratory of Pure Mathematics and Mathematical Practice, East

China Normal University, Shanghai 200241, People’s Republic of China

(YC) School of Mathematical Sciences, Shanghai Key Laboratory of Pure Mathematics and Mathematical Practice, East

China Normal University, Shanghai 200241, People’s Republic of China

(YC) College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao 266590,

People’s Republic of China

(YC) Department of Physics, Zhejiang Normal University, Jinhua 321004, People’s Republic of China

Email address: [email protected]

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